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EARTH  PRESSURE, 
RETAINING  WALLS  and  BINS 


BY 

WILLIAM  CAIN 

Professor  of  Mathematics,  University  of  North  Carolina 
Mem.  Am.  Soc.  C.  E. 


FIRST    EDITION 
FIRST  THOUSAND 


NEW  YORK 
JOHN  WILEY  &  SONS,  INC. 

LONDON:    CHAPMAN    &    HALL,    LIMITED 
1916 


TA  7 


JUbrary 


Copyright,  1916,  by 
WILLIAM  CAIN 


PUBLISHERS  PRINTING  COMPANY 
207-217  West  Twenty-fifth  Street,  New  York 


PREFACE 

OVER  a  century  ago,  Coulomb  formulated  the  laws  of  friction 
and  cohesion  as  affecting  a  mass  of  earth,  and  devised  the"  sliding- 
wedge"  hypothesis  to  effect  the  computation  of  earth  thrust 
against  a  wall.  For  some  reason — doubtless  on  account  of  the 
complexity  of  the  analysis  and  lack  of  experimental  determina- 
tion of  the  coefficients  of  cohesion — the  theory  of  earth  pressure 
was  subsequently  developed  by  many  noted  authors  (Poncelet, 
Weyrauch,  and  others)  after  Coulomb's  hypothesis,  but  for  an 
earth  supposed  to  be  devoid  of  cohesion. 

In  1856,  Rankine  published  his  notable  theory  of  earth 
pressure,  deriving  it  from  considerations  pertaining  to  the 
equilibrium  of  an  infinitesimal  wedge  of  earth  in  the  interior  of 
a  mass  of  homogeneous  earth,  supposed  to  have  a  free  plane 
surface.  Again,  the  earth  was  supposed  to  be  devoid  of  cohesion 
and  likewise  to  be  subjected  to  no  other  external  force  but  its 
own  weight. 

All  of  these  theories  strictly  pertain  to  such  materials  as  dry 
sand,  clean  gravel,  or  loose  rock,  which  are  practically  devoid  of 
cohesion  and  affected  only  by  friction  between  the  particles. 
Although  ordinary  earth  in  bank  is  endowed  with  both  cohesion 
and  friction,  it  was  assumed,  when  this  earth  was  excavated, 
more  or  less  pulverized,  and  placed  behind  a  retaining  wall,  that 
the  cohesion  was  temporarily  destroyed,  so  that  the  theory 
of  earth  endowed  with  friction  alone — considering  the  angle  of 
friction  as  the  angle  of  repose  for  the  loose  earth — could  be  safely 
employed;  since  the  filling,  under  the  influence  of  rains,  settle- 
ment, and  cohesive  and  chemical  affinities,  would  regain  by 
degrees  a  large  part  of  the  cohesion  temporarily  lost,  so  that  the 
thrust  would  ultimately  be  less  than  for  the  pulverized  earth. 
This,  of  course,  tacitly  assumes  that  the  coefficient  of  friction 
would  not  be  lowered  during  the  consolidation. 


4337T3 


iv  PREFACE 

For  many  years,  engineers  have  expressed  their  dissatisfac- 
tion with  a  theory  thus  restricted  and  which,  when  applied  to 
earth  more  or  less  consolidated — especially  clay — was  so  de- 
ficient "in  the  most  vital  elements  existent  in  fact."  It  was 
thus  natural  that  the  pendulum  should  swing  back,  so  that,  in 
very  recent  years,  the  treatment  of  pressures  in  coherent  earth 
has  been  based  on  Coulomb's  original  laws.  Thus  in  ResaPs 
comprehensive  work,*  the  subject  is  treated  analytically  in  great 
generality. 

The  author  was  also  led,  in  the  course  of  a  discussion  of  cer- 
tain experiments  on  retaining  boards  backed  by  earth  and  an 
analysis  of  the  pressures  exerted  on  the  bracing  of  trenches,  to 
develop  a  complete  graphical  method  for  finding  the'  pressures 
in  coherent  earth. f  More  recently,  Mr.  A.  L.  Bell,  M.  Inst. 
C.  E.,  has  added  to  pur  knowledge  of  the  subject  by  experiments 
on  clays  and  an  analysis  concerning  the  supporting  power  of 
foundations,  t 

From  all  of  the  experiments  that  have  been  made  (as  given 
in  Chapter  I),  the  laws  of  Coulomb  seem  to  be  approximately 
verified,  but  it  is  evident  that  extensive  experimenting  upon 
every  kind  of  earth  is  needed  to  give  confidence.  Partly 
from  this  lack  of  experimental  data,  though  mainly  because 
the  theory  of  earth  devoid  of  cohesion  is  strictly  applicable  to  a 
granular  material,  as  clean,  dry  sand,  gravel  or  rip-rap,  the 
theory  for  such  earth  is  fully  developed  in  Chapters  II  and  III, 
and  numerous  applications  are  made  in  Chapter  IV  to  the  design 
of  retaining  walls  of  stone  or  reinforced  concrete. 

It  will  be  found  that  the  analysis  of  Chapters  II-IV  is  more 
critical  and  extended  than  usual.  In  Chapter  *£^  the  discussion 
by  the  "ellipse  of  stress"  method  leads  up  to  Mohr's  "circular 
diagram  of  stress,"  which  is  afterward  used  in  Chapter  V  in 

*  Poussee  des  Terres,  II,  Theorie  des  Terres  Coherentes. 

t  "EXPERIMENTS  ON  RETAINING  WALLS  AND  PRESSURES  ON  TUNNELS,"  by 
Wm.  Cain,  Trans.  Am.  Sac.  C.  E.,  vol.  LXXII  (1911). 

t  Bell  on  "THE  LATERAL  PRESSURE  AND  RESISTANCE  OF  CLAY  AND  THE 
SUPPORTING  POWER  OF  CLAY  FOUNDATIONS."  Proc.  Inst.  C.  E.,  vol.  CXCIX 
(1914-15),  Part  i. 


PREFACE  V 

treating  coherent  earth.  The  author  first  developed  the  theory 
for  coherent  earth  by  the  analytical  method,  but  eventually 
decided  to  use  the  Mohr  diagram,  because  it  not  only  led  to  the 
same  results,  but  gave  numerical  values  with  much  greater  facility 
than  the  formulas  for  the  general  case,  where  the  earth  surface 
is  inclined. 

In  this  chapter,  the  subjects  of  earth  pressures  in  coherent 
earth,  surfaces  of  rupture,  stable  slopes,  foundations,  the  thrust 
against  a  retaining  wall,  the  bracing  of  trenches,  and  the  press- 
ures on  tunnel  linings,  are  treated;  besides,  there  is  added  an 
independent  graphical  method  for  evaluating  earth  thrust. 

The  theory  of  deep  bins  is  given  in  Chapter  VI,  and  the 
attempt  is  made  there  to  reach  fairly  good  results  in  the  vexed 
subject  of  the  thrusts  on  the  walls  of  shallow  bins  filled  with 
coal. 

The  case  of  stresses  in  wedge-shaped  reinforced-concrete  beams 
finds  an  approximate  solution  in  Appendix  I,  in  which  a  number 
of  diagrams  are  added  to  facilitate  computation. 

Finally,  in  Appendix  II,  the  results  of  certain  experiments  on 
model  retaining  walls  are  added,  the  discussion  of  which  may 
prove  instructive. 

It  must  be  borne  in  mind  that  the  theory  of  earth  pressures 
has  been  necessarily  developed  for  a  supposed  homogeneous 
earth,  so  that  it  is  understood  that  its  indications,  for  an  actual 
earth,  must  always  be  supplemented  by  the  practical  judgment 
of  the  experienced  engineer. 

WM.  CAIN. 

CHAPEL  HILL,  N.C. 
Feb.  6,  1916. 


TABLE  OF  CONTENTS 
CHAPTER  I 

LAWS  OF  FRICTION  AND  COHESION.    TABLES,  DIRECTION,  AND 

DISTRIBUTION  OF  STRESS 

ART.  PAGE 

1.  Friction  and  Cohesion  in  Earth I 

2.  Laws  of  Friction  and  Cohesion 3 

3.  Experimental  Method 4 

4.  Earth  Endowed  Only  with  Friction 5 

5.  Angle  of  Repose,  Rankine's  Law  as  to  the  Stability  of  a  Granular 

Mass rl  6 

6-8.  Coefficients  of  Friction  and  Cohesion.     Tables 7 

9.  Weight  of  Earth  in  Water 10 

10.  Exceptional  Case  of  Earth  Thrust 13 

11.  Direction  of  Pressure 14 

12.  Direction  of  Pressure  Against  a  Wall 16 

13.  Direction  of  Pressure  from  Experiments 18 

14.  Direction  of  Pressure  for  Stable  Walls.     Factors  of  Safety        .  19 

15.  Distribution  of  Stress  on  Base 21 

1 6.  Factors  of  Safety  Against  Overturning  and  Sliding    .....  25 

17.  Middle  Third  Requirement,  Etc 26 

CHAPTER  II 

THRUSTS  OF  NON-COHERENT  EARTH.    GRAPHICAL  METHODS 

18.  Surface  of  Rupture * ....  27 

19.  Sliding  Wedge  Theory 28 

20.  Active  and  Passive  Thrust 29 

21.  Graphical  Determination  of  Active  Earth  Thrust  Against  a  Wall  30 

22.  Variation  of  E  with  #' 33 

23.  Examples :,  .  33 

24.  Center  of  Pressure 34 

25.  Definition  and  Use  of  K  and  K\ 35 

26.  Thrust  on  a  Vertical  Plane  in  an  Unlimited  Mass   ...-.«.  35 

27.  Limiting  Plane,  X  <  <f>' 38 

28.  Wall  Above  Limiting  Plane 39 

29.  Wall  Below  Limiting  Plane,  X  >  </>' 39 

30.  Summary    .                   .,..'.....  41 

31.  Tables  for  K,  /3  and  y        ..............  41 

32.  Construction  for  Thrust  on  a  Surcharged  Wall     ......  43 

vii 


CONTENTS 


APPENDIX*   I 

STRESSES  IN  WEDGE-SHAPED  REINFORCED  CONCRETE  BEAMS 

ART.  PAGE 

1-6.  Introductory 239 

7-10.  Stresses  Due  to  Bending;  General  Solution 244 

lO-n.  Applications  to  Counterfort  and  to  Heel 251 

12.  j8  =  o.     Bars  in  One  Plane 253 

13.  ft  >  o.     Bars  in  One  Horizontal  Plane   ........  256 

14.  Prismatic  Beams 258 

15.  Comparison  of  Stresses  in  Concrete  by  Two  Methods       .      .      .  260 

16.  Shear  at  Neutral  Axis 262 

17-18.  Bond  Stress 265 

19.  Variation  in  Shear-Over  Section 268 

20.  Spacing  of  Bars 270 

21.  Compressive   Stresses  in   Concrete   Due   to   Bent    Bars   Under 

Tension 271 

22.  Length  of  Embedment  of  Bars 272 

23.  Working  Stresses  Recommended 273 

APPENDIX  II 

DISCUSSION    OF    EXPERIMENTS    ON    MODEL    RETAINING 

WALLS .     .  276 


Earth  Pressure, 
Retaining  Walls  and  Bins 


CHAPTER  I 

LAWS    OF    FRICTION    AND    COHESION.      TABLES,    DIRECTION,    AND 
DISTRIBUTION   OF   STRESS 

i.  ORDINARY  earths  generally  consist  of  sand  or  gravel  with 
an  admixture  of  clay  or  clayey  matter  and  humus.  When  the 
clayey  matter  is  absent,  we  have  nearly  clean  sand  or  gravel, 
endowed  with  friction  but  nearly  devoid  of  cohesion.  When 
the  sand  or  gravel  is  absent,  we  have  nearly  pure  clay  or  clayey 
earth,  still  possessing  friction  and,  in  addition,  a  large  amount  of 
cohesion.  Between  these  extremes  there  exists  every  kind  of 
earth,  often  saturated  with  water.  Thus  in  bank  sand  or  gravel, 
which  often  includes  a  certain  amount  of  clay,  there  is  a  small 
though  appreciable  amount  of  cohesion,  and  in  ordinary  earths 
before  excavation  a  much  larger  amount. 

Such  earths  will  stand  for  certain  heights  with  a  vertical  face, 
as  is  illustrated  in  the  case  of  trenches,  where  unsupported 
depths  as  great  as  25  feet  have  been  recorded.  If  the  purpose 
in  view  is  to  ascertain  such  heights  or  to  deal  with  the  bracing 
of  trenches  or  tunnel  linings,  then  cohesion  is  a  vital  element  in 
the  computation. 

However,  if  the  design  of  a  retaining  wall  is  in  question,  it  is 
wise  to  ignore  the  cohesion  of  such  earths;  for  when  excavated, 
more  or  less  pulverized  and  deposited  behind  the  wall,  such 
"fresh"  earths  are  temporarily  nearly  devoid  of  cohesion.  It  is 
true  that  as  soon  as  this  fresh  earth  is  deposited,  cohesive  and 
chemical  affinities  come  into  play,  which,  under  the  influence  of 

1 


4  LAWS   OF   FRICTION   AND   COHESION 

cording  to  laws  i  and  2,  are  exerted  at  the  same  time,  on  the 
plane  considered;  so  that  the  question  is  still  an  open  one. 
A  great  deal  of  very  nice  experimenting  will  have  to  be  under- 
taken to  settle  it  finally. 

It  is  probable,  in  connection  with  law  2,  for  very  compressible  substances, 
as  fresh  earth,  that  the  coefficient  of  cohesion  will  increase  with  the  normal 
pressure;  for  the  area  of  actual  contact  of  the  particles  increases  with  the 
pressure,  since  such  pressure  can  squeeze  the  particles  together  and  cause  a 
more  intimate  contact.  Therefore  since  the  cohesion  varies  directly  as  the 
area  of  contact,  it  should  be  greater,  the  greater  the  normal  pressure.  As  an 
illustration,  consider  fresh  pulverized  earth,  whose  coefficient  of  cohesion  is 
small,  though  it  increases  slowly  with  time,  under  the  influence  of  rains,  loads, 
etc. ;  but  take  this  fresh  earth,  wet  it  thoroughly  and  ram  it,  and  it  makes  an 
excellent  puddle  wall,  or  a  cut-off  wall,  to  keep  out  water  under  very  con- 
siderable pressure.  Here  the  cohesion  has  enormously  increased  because  the 
contact  of  the  particles  is  now  so  much  more  intimate  than  when  the  earth  was 
in  a  pulverulent  state. 

As  a  corollary,  it  looks  reasonable  to  suppose  that  the  cohesion  in  a  bank 
of  earth  increases  somewhat  with  its  depth.  If  so,  an  average  value  will  have 
to  be  taken  of  the  coefficient  of  cohesion  for  the  depth  supposed,  for  the 
imaginary  homogeneous  earth,  to  which  the  theory  pertaining  to  coherent 
earth,  as  given  in  Chapter  V,  alone  applies.  It  seems  highly  desirable  then, 
in  experiments,  to  subject  the  earth  to  pressures  corresponding  to  what  it 
actually  sustains  in  banks,  say  from  o  to  50  feet  in  height,  to  ascertain  the 
variation,  if  any,  in  the  coefficient  of  cohesion. 

3.  A  simple  way  to  find  experimentally  the  coefficients 
/  and  c  can  be  illustrated  by  Fig.  i ,  where  a  thin  slice  of  earth  is 


FIG.  i 


supposed  to  be  placed  between  two  rough  metallic  plaques, 
which  are  then  to  be  firmly  pressed  together  without  contact 
and  the  resistance  to  the  relative  displacement  of  the  two  plaques 


3,  4]  EARTH   DEVOID    OF   COHESION  5 

recorded.  Suppose  the  pressure  to  be  due  to  a  weight  and  let 
W  =  weight  of  earth  and  plaque  above  the  plane  of  shear  AB 
•f  any  load  resting  on  the  upper  plaque;  also  let  P  be  such  a 
force,  applied  to  the  upper  plaque  in  the  plane  AB,  that  motion  or 
sliding  of  the  earth  above  the  plane  AB  over  the  earth  below  it 
is  impending.  This  is  taken  to  mean,  throughout  the  book, 
that  no  motion  ensues,  but  that  the  slightest  increase  in  P  will 
cause  motion.  The  earth  along  the  plane  AB  resists  the  force 
P  by  its  friction  and  cohesion,  acting  to  the  left  along  AB.  The 
normal  reaction  ^V  of  the  earth  below  AB  is  equal  to  W]  hence 
by  law  i  the  frictional  resistance  =  fN  =  fW.  If  now,  the 
area  of  the  plane  AB  on  which  shearing  is  impending  is  called  a, 
the  cohesive  resistance,  by  law  2  is  ca. 
Hence  for  equilibrium, 

P=fN  +  ca=fW  +  ca     ....     (i) 

When  several  experimental  determinations  of  P  for  different 
values  of  W  have  been  made,  we  can  form  a  series  of  equations, 
of  the  form  (i),  and  by  elimination  deduce  the  corresponding 
values  of  /  and  c. 

In  Fig.  i,  if  we  lay  off  vertically  downward,  ON\  =  W,  and 
horizontally  to.  the  right,  N\S\  =  P,  then  OSi  =  resultant  of  P 
and  W.  Also  if  we  lay  off  vertically  upward  ON  =  N  =  W 
and  from  N,  horizontally  to  the  left,  NR  =  fN=fW  and  R  S 
=  ca,  then  by  eq.  (i),  NS  =  fN  +  ca  =  P  and  0  S  the  resultant 
of  N  and  (fN  +  ca)t  is  equal  and  directly  opposed  to  OSi,  the 
resultant  of  P  and  W.  If  P  is  <  JW  +  ca,  then  only  so  much  of 
the  friction  and  cohesion  as  is  needed  to  equilibrate  P  is  exerted* 

4.  Earth  Devoid  of  Cohesion.  For  non-coherent  earth, 
c  =  o  and  R  S  =  o,  whence  for  equilibrium  P  =  NiRi  and  0  RI, 
the  resultant  of  P  and  W,  is  equal  and  directly  opposed  to  0  R, 
the  resultant  of  N  and  fN. 

The  angle  NOR  =  <p  is  called  the  angle  of  friction.      We  have, 

NR    fN  . 

tan<?  =  —  =  -==/ (2) 

Hence  for  the  impending  motion,  the  resultant  reaction  O'R 
always  makes  the  angle  <p  =  tan~  /,  with  the  normal  reaction  of 


6  LAWS   OF  FRICTION  AND   COHESION 

the  plane  AB.    Also, 

P  =  NiRi  =  NR=fN  =fW  =  Wtan<p. 

The  same  principles  and  formulas  apply,  when  one  solid  body 
is  on  the  point  of  sliding  over  another,  the  surfaces  of  contact 
being  plane. 

5.  If  a  block  is  placed  on  an  inclined  plane,  Fig.  2,  making  an 
angle  a  with  the  horizontal,  and  slipping  downward  is  impend- 
ing, the    total    reaction    of    the    plane 
OR  must  be  equal  and  opposed  to  the 
weight  W  of  the  block  which  acts  ver- 
tically downward.      Hence,  when  cohe- 
sion   is  ignored,   since   OR  makes   the 
angle  <p  —  NOR,  with  the  normal  to  the 
plane,  it  follows  that  <p  =  a,  the  sides  of 
the  two  angles  being  perpendicular,  each 

to  each.  The  angle  a  is  called  the  angle  of  repose  for  the  two  sur- 
faces in  contact.  Since  a  =  <p,  it  is  easy  to  find  the  coefficient  of 
friction  between  any  two  bodies  by  placing  one  on  the  other  and 
tilting  the  second  one  until  motion  just  ensues.  On  measuring 
the  corresponding  angle  a,  we  have/  =  tan  <p  =  tan  a. 

If  the  two  bodies  are  both  earth,  the  plane  inclined  at  the 
angle  <p  (the  angle  of  friction  of  earth  on  earth)  to  the  horizontal 
is  the  plane  of  steepest  slope  on  which  the  earth  particles  will 
not  slide,  and  the  angle  <p  is  sometimes  called  the  angle  of  natural 
slope. 

As  a  corollary  to  what  precedes,  the  following  theorem  for 
non-coherent  earth,  due  to  Rankine,  may  be  stated: 

THEOREM.  //  is  necessary  to  the  stability  of  a  ^nular  mass, 
that  the  direction  of  the  pressure  between  the  portions^ihw  which  it  is 
divided  by  any  plane  should  not  at  any  point  make  with  the  normal 
to  that  plane  an  angle  exceeding  the  angle  of  repose. 

6.  The  coefficients  of  friction  for  a  great  number  of  bodies 
have  been  found  experimentally,  generally  after  the  method  first 
given.     Thus,  as  in  Fig.  i,  the  force  P  is  found  that  will  just 
start  one  body  sliding  on  another.     If  the  weight  of  the  upper 
body,  with  any  load  that  may  be  placed  on  it,  is  W,  then,  /  = 


6,7]  COEFFICIENTS   OF  FRICTION  AND   COHESION  7 

tan  (p  =  P/W.  This  ignores  any  cohesion.  If  there  is  cohesion, 
then  this  method  of  procedure  gives  too  large  a  value  for  /  = 
tan  <p,  in  that  it  substitutes  NS  for  NR  in  Fig.  i ,  or  it  assumes  <p 
=  NOS  in  place  of  the  true  value  NOR.  Thus  the  coefficient  of 
" static  friction"  is  always  too  large.  When  motion  once  begins, 
then  presumably  all  cohesion  is  destroyed,  and  it  would  seem 
that  the  so-called  coefficient  of  kinetic  friction  is  nearer  the  true 
value  of/. 

7.  Coefficients  of  Friction  and  Cohesion.  Very  few  ex- 
periments have  been  made  with  the  object  of  determining  the 
coefficients  of  cohesion. 

In  the  experiments  of  Leygue,*  an  inclined  plane  was  used, 
but  the  laws  of  Coulomb,  as  symbolized  by  eq.  (i),  were  assumed 
as  the  basis  of  the  computation.  In  these  experiments,  the 
normal  pressures  were  very  small — only  7  to  30  lb./ft.2 — but  the 
laws  of  Coulomb  were  verified  within  these  narrow  limits.  The 
coefficients  of  friction/,  with  the  corresponding  angle  of  friction 
<p,  together  with  the  coefficient  of  cohesion  c  in  lb./ft.,2  as  com- 
puted, are  as  follows: 

Dry  sand,  /  =  0.70,  <p  =  35°       ,  c  =    1.47 

Wet  sand,  0.85,          40°  22',  8.28 

Very  wet  sand,  1.70,          59°  30',  6.36 

Damp  fresh  earth,          1.63,          58°  28',          18.45 
Leygue  states  that  Collin  found,  by  an  independent  method, 
for  clayey  earth,  c  =  23.1,  and  for  clay  of  little  consistency, 
c  =  39.5  lb./ft.2 

The  next  series  of  experiments  were  made  in  1910  by  Mm. 
Jacquinot  and  Frontard,  on  earth  taken  from  an  earthen  reservoir 
dam  that  had  failed  from  a  considerable  lowering  of  the  water 
level. f  To  the  earth,  water  was  added,  and  it  was  then  worked 
or  kneaded  and  compressed  with  the  hands  so  as  to  make  a  firm, 
though  pasty  cake,  which  was  then  inserted  between  the  plaques, 
Fig.  i.  The  pressures  were  high  but  the  laws  of  Coulomb  were 
again  approximately  verified.  The  results  are  as  follows: 

*  Annales  des  Fonts  et  Chaussees,  Nov.,  1885. 

t  Given  in  some  detail  in  Resal,  "Poussee  des  Terres,"  Deuxieme  Partie, 
P-  327- 


8  LAWS   OF   FRICTION  AND    COHESION 

Normal  pressure  Coeff.  friction.  Coeff.  cohesion. 

N  f  =  tan  <p  c 

lb./ft.2  <p  =  8°  lb./ft.2 

692  /  =  0.14  395 

2980  420 

5665  "  403 

7*54  448 

The  earth  weighed  112  lb./ft.3,  as  taken  from  the  bank. 
The  pressures  recorded  would  correspond  to  the  vertical  press- 
ures experienced  at  depths  of  6  to  64  feet  in  a  bank  of  the 
earth  considered.* 

The  large  values  of  c  were  to  be  expected,  since  the  theory  of 
open  cuts  or  trenches  would  lead  us  to  expect  for  ordinary  con- 
solidated earth  values  of  c  running  into  several  hundred  pounds; 
but  the  very  small  value  of  /  ==  tan  <p  is  startling,  for  it  means 
that  we  may  have  to  revise  our  ideas  very  greatly  with  reference 
to  this  coefficient  of  friction  for  even  ordinary  earth. 

It  may  be  noted,  in  this  connection,  that  certain  earths  can 
exert  an  appreciable  amount  of  tension.  Thus  Mr.  W.  Airyf 
states  that  he  found  the  tensile  strength  of  a  block  of  ordinary 
brick  clay  to  be  168  and  of  a  certain  shaley  clay  800  lb./ft.2. 
Since  the  maximum  intensity  of  shear  in  a  prismatic  block  is  one- 
half  the  unit  tension,  it  is  seen,  for  these  two  specimens,  that  the 
unit  shearing  or  cohesive  strength  must  have  been,  at  least, 
84  and  400  lb./ft.2  respectively. 

Experiments  have  also  been  made  by  Mr.  Arthur  Langtrv 
Bell,  M.  Inst.  C.  E.,  on  certain  clays  encountered  in  sinking  a 
monolith  at  Rosyth,  to  determine  the  coefficients  of  friction  and 


*  There  were  other  experiments  performed  upon  the  earth  as  it  was  taken 
from  the  dam,  also  after  it  had  been  rammed,  but  the  manometers  were  out 
of  order  and  the  results  uncertain.  It  was  thought,  though,  that  the  following 
conclusions  could  be  fairly  drawn: 

(1)  /  =  tan  <p  varied  only  from  0.14  for  the  soft,  pasty  earth  to  0.18  for 
earth  nearly  dry.     Neither  amount  of  water  used  nor  the  puddling  or  ramming 
caused  much  variation  in  /. 

(2)  On  the  contrary,  the  amount  of  water  used  affects  the  cohesion  very 
much,  and  sufficient  ramming  can  more  than  double  the  coefficient  of  cohesion. 
If  is  evident  that  a  great  deal  of  experimental  work  along   these   lines  should 
be  done  to  reach  safe  and  practical  conclusions. 

t  Minutes  of  Proceedings,  Inst.  C.  E.,  vol.  LXV,  p.  140. 


7,8] 


TABLES  OF  ANGLE  OF  REPOSE 


3" 

5° 

7° 

16° 


cohesion.*     The  following  average  values  of  c  and  $  are  given 
by  Mr.  Bell:  c 

Tons  per  ^ 

Sq.  Ft. 

Very  soft  puddle  clay 0.2  ....     o° 

Soft  puddle  clay 0.3  . 

Moderately  firm  clay 0.5  . 

Stiff  clay 0.7  . 

Very  stiff  boulder  clay 1.6  . 

By  " puddle  clay"  is  meant  a  pure,  homogeneous,  plastic 
clay,  free  from  sand  and  stones.  The  .author  states  that  for 
"perfectly  dry  sand,"  c  =  o  and  <p  =  angle  of  repose,  if  the  sand  is 
rammed  as  it  is  inserted  in  the  cylinder;  but  if  merely  poured  in 
and  shaken,  the  experimental  value  of  <p  was  much  less  than  the 
angle  of  repose.  For  wet  sand  grabbed  from  a  monolith  well,  the 
values  found  were:  c  =  0.3  ton  per  sq.  ft.,  <p  =  31°. 

8.  Tables  of  Angles  of  Repose,  Weights,  etc.  The 
following  table  gives,  for  loose  earth  or  other  granular  material, 
the  angle  of  repose,  or  the  steepest  inclination  of  the  surface  to 
the  horizon,  and  the  weights  in  pounds  per  cubic  foot. 

TABLE  I 
ANGLES  OF  REPOSE  AND  WEIGHTS  PER  CUBIC  FOOT 


Material 


Angle  of 
Repose 


Weight  Lb. 
per  Cu.  Ft. 


Sand,  dry.  .  .  .  . 

Sand,  moist 

Sand,  wet 

Ordinary  earth, f  dry 

Ordinary  earth,  moist 25 

Ordinary  earth,  wet 25 

Gravel,  round  to  angular :•    30 

Gravel,  sand  and  clay r-    20 

Cinders,  bituminous 25 

Coke 30 

Coal,  anthracite 

Coal,  bituminous 

Soft,  flowing  mud 

Soft,  rotten  rock 

Hard,  rotten  rock 

Ashes,  anthracite 

Rip-rap , 


to  35 
to  45 
to  40 

to  45 
to  45 
to  30 
to  48 

to  37 
to  41 

to  45 

27 

35 
o 

37 

45 

45 

'  45 


90  to 

100  to 

no  to 

80  to 

80  to 

100  to 

100  to 

100  tO 

45 

23  to 
40  to 
44  to 

105  to 

no 

100 

no 


no 
no 

120 
100* 
100 
120 
135 


32 

55 
54 

120 


*  Min.  Proc.  Inst.  C.  E.,  vol.  CXCIX,  session  1914-15.     Part  i. 
t  A  loose  mixture  of  sand  or  gravel  with  clay  and  perhaps  humus — loam. 
With  20%  clay,  it  may  be  called  a  sandy  loam;  with  50%  clay,  a  clay  loam. 


10 


LAWS  OF  FRICTION  AND  COHESION 


The  foregoing  table,  as  well  as  the  tables  that  follow,  have 
been  compiled  mainly  from  "American  Civil  Engineers'  Pocket 
Book,"-"Trautwine's  Engineers'  Pocket  Book,"  and  Rankine's 
Treatises. 

TABLE  II 
COEFFICIENTS  AND  ANGLES  OF  FRICTION 


^ 

/ 

* 

Masonry  upon  masonry  .  

o  6s 

**• 

Masonry  upon  wood,  with  grain  

o  60 

31° 

JMasonry  upon  wood  across  grain 

o  so 

26°  40' 

Masonry  on  dry  clay 

o  50 

26°  40' 

Masonry  on  wet  clay    

O    "\^t 

1  8°  20' 

Masonry  on  sand  

o  4.0 

21°  SO' 

Masonry  on  gravel  

o  60 

3  ° 

Dry  bituminous  coal  on  iron 

o  84 

40° 

Dry  bituminous  coal  on  wood 

I    OO 

*§**„ 

AS 

Dry  anthracite  coal  on  iron  

o  s8 

30° 

Dry  anthracite  coal  on  wood  

O   7O 

35° 

9.  Weight  of  Granular  Material  in  Water.  Suppose  the 
rilling  behind  a  retaining  wall  to  consist  of  approximate  spheres, 
like  marbles,  .  permitting  free  access  to  the  water  covering 

TABLE  III 
WEIGHTS  PER  CUBIC  FOOT  OF  MASONRY 


Kind  of  Masonry 

Weight  Lb. 
per  Cu.  Ft. 

Cinder  concrete  

IIO 

Stone  concrete  

140  to  150 

Reinforced  stone  concrete 

ISO 

Brick  masonry,  soft  

IOO 

Brick  masonry,  common  

I2S 

Brick  masonry,  pressed  

1  4.O 

Dry  rubble   sandstone 

IIO 

Dry  rubble,  limestone  '.  

I2S 

Dry  rubble,  granite  #•»!. 

» 
|    I-IQ 

Mortar  rubble,  sandstone  

no 

Mortar  rubble   limestone 

ISO 

Mortar  rubble,  granite  . 

ISS 

Ashlar,  sandstone  

I4.O 

Ashlar,  limestone  

1  60 

Ashlar,  granite 

i6s 

it,  on  all  sides.     If  this  filling  weighs  100  Ibs.  per  cu.  ft.  in  air 
and  has  40  per  cent  voids,  then  i  cu.  ft.  of  the  filling  contains 


9] 


WEIGHT  IN  WATER 


11 


0.6  cu.  ft.  of  solids,  and  since  the  buoyant  force  of  the  water 
equals  the  weight  of  an  equal  volume  of  water  or  0.6  X  62.5 
=  37.5  Ibs.,  the  true  weight  of  the  filling  in  fresh  water  is  100 
—  37.5  =  62.5  Ibs.  per  cu.  ft.* 


TABLE  IV 
SAFE-BEARING  CAPACITY  OF  SOILS 


Soil 

Tons  (of 
2,000  Lbs.) 
per  Sq.  Ft. 

Quicksand,  alluvial  soils  

o  c  to       i 

Clay,  soft  

o  5  to      2 

Clay  and  sand  

2      to      4 

Sand,  ciean,  dry  

2        tO        4 

Sand,  compact,  well  cemented  

,                    U"             T-       j 

4      to      6  ' 

Gravel  and  coarse  sand  

6      to      8 

Ditto,  well  cemented 

8      to    10 

Clay,  hard,  moderately  dry 

4      to      6 

Clay,  hard,  dry 

6      to      8 

Rock,  soft  to  hard    . 

5      to  200 

This  method  can  be  applied  to  get  the  approximate  weights 
in  water  of  such  materials  as  rip-rap,  cobble-stones,  or  clean 
gravel.  It  cannot  apply  to  the  finer  materials,  as  clean  sand, 
bank  sand,  or  loam,  because  in  such  earths  free  circulation  of  the 
water  is  not  realized;  the  water  cannot  get  under  the  material 
everywhere  and  its  full  buoyant  effect  cannot  be  exerted.  How- 
ever, from  lack  of  sufficient  experimental  data,  the  method  is 
applied  to  all  earths,  so  that  if  the  filling  weighs  in  air  w  Ibs.  per 
cu.  ft.  and  i  cu.  ft.  contains  n  cu.  ft.  of  solids,  then,  reasoning  as 
above,  the  weight  per  cu.  ft.  of  the  filling  in  fresh  water  is, 

(w  —  n  X  62.5)  Ibs. 
If  p  =  per  cent  of  voids,  n  =  (i  —p/ioo). 

With  this  weight  of  filling  in  water,  the  earth  thrust  is  com- 
puted, for  the  angle  of  repose  of  the  earth  in  water  and  the  water 
pressure  against  the  wall  will  be  estimated  for  the  full  head 
of  water.  For  salt  water,  replace  62.5  in  formula  above  by  64. 


*  It  is  perhaps  well  to  caution  here  against  the  very  common  error  of 
subtracting  62.5  from  the  weight  per  cu.  ft.  of  the  filling  in  air  to  get  its  weight 
per  cu. 


ft.  in  fresh  water. 


12 


LAWS   OF  FRICTION  AND   COHESION 


An  interesting  experiment  on  the  transmission  of  pressure  through  satu- 
rated sand  has  been  described  by  Mr.  J.  C.  Meem,*  in  which  it  was  shown 
that  the  water  pressure  on  a  given  area,  through  6  or  8  inches  of  sand,  with 
40%  voids,  was  about  40%  of  the  static  head.  The  pressure  averaged  4.5 
Ibs.  per  sq.  in. 

It  will  prove  instructive  to  apply  the  results  to  the  case  of  sand  weighing 
100  Ibs.  per  cu.  ft.  in  air,  with  40%  voids.  The  earth,  in  water,  is  assumed  to 
transmit  only  40%  of  the  water  pressure  due  to  the  full  static  head;  conse- 
quently a  cubic  foot  of  the  earth,  containing  only  0.6  cu.  ft.  of  solids,  will  be 
subjected  to  an  effective  buoyant  force  of  only  0.4  X  0.6  X  62.5  =  15  Ibs.; 
whence  the  earth  will  weigh  100  —  15  =  85  Ibs.,  in  water.  The  earth  thrust 
against  the  retaining  wall  is  now  to  be  computed  for  this  weight  per  cu.  ft. 
and  the  corresponding  angle  of  repose  of  the  earth  in  water.  The  water 
pressure  on  the  wall  from  the  surface  of  the  saturated  earth  downward  will 
then  be  taken  as  only  40%  of  that  due  to  the  static  head. 

It  is  to  be  hoped  that  experiments  on  a  large  scale,  with  varying  pressures 
and  thicknesses  of  sand  and  earth,  will  be  undertaken  to  establish  more 
definitely  the  laws  of  the  transmission  of  water  pressure  through  earth.  In 
this  connection,  see  "The  Action  of  Water  Under  Dams,"  by  J.  B.  T.  Colman, 
in  Proc.  Am,  Soc.  C.  E.,  Aug.,  1915. 

By  kind  permission  of  Prof.  Mansfield  Merriman,  the  follow- 
ing table  is  quoted  from  the  "American  Civil  Engineers'  Pocket 
Book." 

Material  excavated  by  a  wet  or  dry  process  and  dumped 
into  water,  as  at  the  back  of  a  sea  wall,  has  weights  and  slopes 
approximated  as  follows: 

TABLE  V 


Kind  of  Material 

Slope 
of 
Repose 

Angle 
v        of 
Repose 

Weight 
Lb.  per 
Cu.  Ft. 

Sand,  clean  
Sand  and  clay 

2  to 
•J  tO 

26°  34' 
18    26 

60 
6s 

Clay  

•j    c  to 

ic    57 

80 

Gravel,  clean  

2  tO 

26    ^4 

60 

Gravel  and  clay  

^  to 

18    2  -,*  , 

65 

Gravel,  sand  and  clay  
Soil  

3  to 

7     C   tO 

18    26^  J 
It;    57 

65 
7O 

Soft,  rotten  rock  

I   tO 

45    °° 

65 

Hard  rock,  rip-rap 

I   tO 

4.5    oo 

65 

River  mud  .  .  .  

GO   tO 

o   oo 

oo 

The  slopes  given  were  observed,  the  weights  computed  for 
assumed  voids.     In  any  specific  case,  it  is  desirable  for  the 


*  Trans.  Am.  Soc.  C.  E.,  vol.  LXX,  pp.  366-369. 


9,10] 


WEIGHT   IN   WATER 


13 


engineer  to  obtain  the  slope  and  weight  of  the  earth  by  observa- 
tion. Lacking  that,  it  is  common  practice  to  take  the  weight 
of  ordinary  dry  earth  at  100  Ib.  per  cu.  ft.  and  the  angles  of  re- 
pose as  30°  or  33°  41',  the  latter  corresponding  to  the  slope  of  i^ 
base  to  i  rise. 

10.  In  the  following  chapters  methods  will  be  given  for 
estimating  the  earth  thrust  for  homogeneous  earth  against  a 
retaining  wall,  but  it  may  be  well  to  caution  the  engineer 
here  that  exceptional  cases  arise  where  the  ordinary  theory  is 
inadequate  and  which  will  require  the  exercise  of  his  best 
judgment. 

Thus,  certain  clays  swell  when  exposed  to  the  air  with  great 
force,  or  perhaps  the  filling  may  be  deposited  upon  rock  or  clay 
sloping  toward  the  wall  and  water  from  springs  or  rains  lubri- 
cates the  rock  or  clay  base,  lessening  the  frictional  resistance 
and  greatly  increasing  the  thrust.  As  a  simple  illustration,  let 
us  suppose  that  the  rock  base  makes  an  angle  i  with  the  horizon- 
tal, see  Fig.  3,  and  that  in  consequence  of  springs  at  C,  there  is  a 
break  in  the  earth  along  CD  and  that  water  freely  finds  its  way 
along  the  rock,  destroying  any  possible  cohesion  there  and  re- 
ducing the  coefficient  of  friction  of  earth  on  rock  to  /  =  tan  <p. 


FIG.  3 


Call  the  weight  of  the  earth  ABC  that  tends  to  slide  on  the 
rock  base  AD,  for  one  foot  length  of  wall,  W,  the  reaction  of  the 
wall  AB,  E,  and  its  horizontal  component  EI.  Then  W  sin  i  = 
component  of  W  in  direction  DA ,  and  the  normal  component  is 
W  cos  i;  whence  the  frictional  resistance  acting  up  the  plane  or 
in  the  direction  AD  is  W  cos  i  tan  <?,  by  Coulomb's  law.  The 


14  LAWS   OF  FRICTION  AND   COHESION 

force  tending  to  cause  the  mass  BADC  to  slide  down  AD  is  thus, 
W  sin  i  —  W  cos  i  tan  <? 

and  for  equilibrium  of  this  mass,  its  horizontal  component  must 
equal  EI. 

W  sin  (i  —  <p) 
.'.  EI  =  W  (sin  i  —  cos  i  tan  <p\  cos  i  = cos  i. 

COS  <p 

The  direction  of  E  will  receive  attention  in  the  next  article. 
The  value  found  for  its  horizontal  component  is  independent  of 
that  direction. 

ii.  Direction  of  Pressure.  Consider 
an  unlimited  mass  of  earth  with  a  plane 
upper  surface  AB,  Fig.  4,  making  the  angle 
i  with  the  horizontal.  The  plane  of  the 
paper  in  Fig.  4  represents  a  vertical  sec- 
tion cutting  the  surface  in  the  line  AB 
of  greatest  slope  and  BCD  a  vertical  plane 
perpendicular  to  the  plane  of  the  paper. 
Consider  the  conditions  of  equilibrium  of  the  prism  CDEF, 
bounded  by  planes  parallel  to  the  surface  and  laterally  by 
vertical  planes,  two  of  them,  CD  and  EF,  being  perpendicular 
to  the  section,  and  two  parallel  to  it,  and  one  unit  apart. 
Let  BC  =  x,  CD  =  A  x  and  regard  the  earth  as  homogeneous  and 
weighing  w  Ib.  per  cu.  ft.;  also  let  the  common  area  of  the  upper 
and  lower  surfaces  of  the  elementary  prism  be  unity.  Thus  if 
FC  =  i  ft.,  the  area  of  the  horizontal  section  of  the  vertical 
prism  BF  is  cos  i  sq.  ft.  and  the  weight  of  this  prism  is  wx  cos  i 
pounds.  //  the  earth  is  exposed  to  no  external  j*Z'Je  but  its  own 
weight,  the  only  pressure  that  the  elementary  prism  CDEF  can 
sustain  on  its  upper  surface  is  the  weight  of  earth  directly  over 
it,  or  w  x  cos  i  Ibs.,  acting  vertically  downward.  Similarly  the 
downward  pressure  on  the  lower  surface  ED  is  (x  +  A  x)  w 
cos  i  and  this  is  equal  and  directly  opposed  to  the  reaction  on 
ED,  acting  vertically  upward.  The  vertical  forces  acting  on 
the  prism  CDEF  (the  weights  of  BF  and  EC  and  the  reaction 
of  ED)  thus  balance  independently.  Hence  the  pressures  on  the 


FIG.  4 


11]  DIRECTION  OF  PRESSURE  15 

vertical  planes  CD,  EF,  must  balance  independently;  conse- 
quently they  must  act  parallel  to  the  sloping  surface  and  be  of 
equal  intensity  at  the  same  depth.  The  pressures  acting  on  the 
vertical  faces  of  the  prism  CDEF,  parallel  to  the  plane  of  the 
paper,  are  evidently  horizontal,  from  considerations  of  sym- 
metry, and  balance  independently. 

It  has  now  been  shown  that  the  pressure  on  a  plane  parallel 
to  the  upper  plane  surface  of  the  earth  is  vertical  and  pro- 
portional to  the  depth;  the  pressure  on  the  vertical  plane  CD  is 
parallel  to  the  upper  surface  and  finally,  the  state  of  stress  at  a 
given  depth,  is  uniform. 

These  are  the  conclusions  of  Rankine,  as  given  in  his  "Ap- 
plied Mechanics,"  Arts.  101,  125,  195,  and  they  are  sound  for 
the  case  supposed,  where  the  earth  is  subjected  to  no  external 
force  but  its  own  weight. 

The  vertical  pressure  on  CF  and  the  pressure  on  CD  are 
said  to  be  conjugate  pressures,  each  acting  parallel  to  the  plane 
on  which  the  other  acts.  Also  each  stress  makes  the  same 
angle  i  with  the  normal  to  the  plane  on  which  it  acts,  or  they 
have  the  same  obliquity,  the  angle  which  any  stress  makes  with 
the  normal  to  the  plane  on  which  it  acts  being  called  its  obliquity. 

The  conclusions  above  hold  if  an  incompressible  earth  is 
deposited,  to  the  same  depth  everywhere,  on  an  incompressible 
plane  foundation,  inclined  or  horizontal.  They  likewise  hold 
for  a  compressible  earth,  even  when  the  foundation  is  com- 
pressible, provided  the  compression  of  the  foundation  is  every- 
where uniform,  and  any  plane  in  the  earth  parallel  to  the  free 
surface  remains  plane  and  parallel  to  the  surface  after  the  com- 
pression of  the  earth,  since  the  conditions  assumed  do  not  affect 
the  reasoning. 

But  if  the  foundation  gives  more  in  some  places  than  others, 
then  the  earth  will  sink  more  where  the  foundation  is  most 
yielding.  This  sinking  is  resisted  to  a  certain  extent  by  the 
friction  resulting  from  the  thrust  of  the  earth  surrounding  the 
falling  mass,  so  that  part  of  its  weight  is  transmitted  to  the 
sides,  as  actually  happens  in  the  case  of  fresh  earth  deposited 
over  drains  or  culverts.  When  a  tunnel  is  driven  through 


16  LAWS   OF  FRICTION  AND   COHESION 

earth,  most  of  the  weight  of  the  material  vertically  over  the 
tunnel  is  transmitted  to  the  sides,  only  a  small  portion  being 
sustained  by  the  tunnel  lining.  Even  when  the  foundation  is 
practically  unyielding,  if  it  is  not  plane,  so  that  the  depth  of 
earth  varies  very  materially,  the  earth  of  greatest  depth  will 
sink  most,  so  that  part  of  its  weight  will  be  transmitted  to  the 
sides  through  friction.  Thus  in  large  masses  of  earth,  as  in 
embankments  as  we  usually  find  them,  the  pressures  are  not 
generally  the  same  at  the  same  depth.  In  fact,  the  conditions 
originally  posited  are  rarely  fulfilled  in  ordinary  practice. 

12.  Direction  of  the  Thrust  Against  a  Retaining  Wall. 
If  we  imagine  an  immovable  rigid  wall  on  an  unyielding 
foundation,  backed  by  an  incompressible  granular  mass  having 
a  plane  free  surface,  then  the  pressure  of  the  filling  on  a  vertical 
plane  entirely  in  the  earth  and  touching  the  wall  will  be  parallel 
to  the  free  surface,  by  the  preceding  theory.  But  in  practice,  the 
wall,  filling,  and  foundation  are  compressible,  and  almost  in- 
variably the  filling  is  much  more  compressible  than  the  wall,  so 
that  it  settles  more  than  the  wall.  As  a  consequence,  the  earth 
rubs  against  the  wall  in  its  relative  downward  motion,  developing 
friction  and  thus  introducing  an  extraneous  force  not  con- 
templated in  the  foregoing  (Rankine)  theory.  Again,  the  wall 
and  foundation  being  compressible,  the  top  of  the  wall  moves 
over  slightly  from  the  earth  thrust,  if  the  resultant  of  this  thrust 
and  the  weight  of  the  wall  passes  outside  the  center  of  the  base 
of  the  wall.  It  is  only  for  certain  batters  of  the  back  of  the 
wall  (to  be  investigated  in  Arts.  27-30),  that  a  certain  wedge 
of  earth  moves  with  the  wall  as  it  moves  over,  in  which  case  the 
earth  sooner  breaks  along  an  interior  plane  iiw^te  mass  than 
along  the  wall.  Excluding  these  cases  for  the  present,  for  all 
other  cases,  as  the  top  of  the  wall  moves  over,  however  slightly, 
or  strictly,  when  such  motion  is  impending,  the  earth  will  get 
a  grip  on  the  wall,  acting  downward  along  the  back  of  the.  wall, 
equal  to  the  true  normal  component  of  the  earth  thrust  mul- 
tiplied by  the  coefficient  of  friction  of  earth  on  masonry,  and  it  is 
impossible  for  the  wall  to  overturn  or  slide  without  this  full 
frictional  resistance  being  exerted.  • 


12]  DIRECTION   OF   THRUST  17 

If  the  top  of  the  wall  leans  toward  the  earth,  this  theory  is 
only  approximately  true.  If  the  back  of  the  leaning  wall  makes 
an  angle  with  the  vertical  not  exceeding  about  10°,  or  for  usual 
inclinations,  the  theory  is  sufficiently  near  for  computation  of 
the  resulting  thrust.  See  Art.  62.  Let  us  apply  this  theory 
to  the  retaining  wall  A  BCD,  Fig.  5, 
backed  by  earth  ABL,  having  an  angle 
of  friction  <p,  the  angle  of  friction  of  earth 
on  wall  being  <p',  all  cohesion  being  neg- 
lected. If  the  true  normal  component  of 
the  earth  thrust  on  AB  is  called  EI,  then 
from  the  tendency  to  relative  settling  of 
the  earth  or  from  the  impending  moving  Fip.  5 

over  of   the    top    of    the    wall    or    from 

both  causes,  the  earth  will  exert  a  friction  force,  acting  always 
opposed  to  the  incipient  motion,  or  along  BA,  in  the  direction 
BA,  of  amount  EI  tan  <f>f  or  EI  tan  <p,  if  <p'  >  <p.  In  the  latter 
case  ((p  ><p),  a  thin  layer  of  earth  will  cling  to  the  wall,  so  that 
in  case  of  the  impending  relative  motion  supposed,  this  layer, 
rubbing  against  the  remaining  earth,  will  only  cause  the  friction 
due  to  earth  on  earth.  Let  us  suppose  <pf  fL  <p,  then  the  fric- 
tion force  is  EI  tan  </.  The  reaction  MI  of  the  wall  can  be 
decomposed"  into  two  components;  one  normal  to  the  wall, 
N  I  =  EI,  since  it  is  equal  and  opposed  to  the  normal  compo- 
nent of  the  earth  thrust;  the  other  01,  equal  and  opposed  to  the 
friction  force  EI  tan  <p  '. 

Completing  the  parallelogram  M  O  I  N,  we  have  M  N  =  0  1 

EI  tan  (p  \  whence, 

M  N      EI  tan  <p 
tan  NIM  =  ~       =  -  —  —      =  tan  <p 


.'.  NIM  =  <p', 

or,  the  reaction  of  the  wall  makes  the  angle  <?'  with  the  normal 
to  the  wall.  When  <?'  >  <p,  as  shown  above,  the  friction  force 
is  EI  tan  <p;  hence  in  this  case  NIM  =  <p. 

It  is  thus  seen  that  the  reaction  of  the  wall,  when  friction  at 
the  back  of  the  wall  is  exerted,  makes  the  angle  <p'  with  the  normal 


18  LAWS   OF   FRICTION  AND   COHESION 

to  the  back  of  the  wall  when  $  <_  <p,  or  the  angle  <p  with  the  nor- 
mal when  <p  >  (p.  Consequently,  when  AB  is  vertical,  the  total 
earth  thrust  on  the  wall,  which  is  equal  and  directly  opposed 
to  MI,  the  reaction  of  the  wall  is  not  parallel  to  the  free  surface, 
as  in  the  Rankine  theory,  save  when  the  free  surface  is  at  the 
angle  of  repose  and  NIM  =  (p.  The  extraneous  force — the  wall 
friction — causes  this  change  in  the  usual  direction  of  the  earth 
thrust  on  a  vertical  plane. 

It  will  be  observed,  in  the  analysis  above,  that  no  actual 
relative  motion  is  required,  but  only  " impending  "relative  motion 
of  earth  and  wall. 

13.  It  will  now  naturally  be  inquired,  What  has  experiment 
to  say  as  to  the  direction  of  the  earth  thrust  against  a  vertical 
retaining  wall  ?  Is  it  parallel  to  the  free  surface  or  does  it  make 
an  angle  <p  or  v  (<?'  >  <p)  with  the  normal  to  the  wall?  In 
answer  to  this  question  in  a  general  way,  let  us  consider  a  very 
light,  empty  wooden  box  of  a  width  slightly  greater  than  the 
height  and  of  any  length  and  backed  by  sand  level  with  the  top 
of  the  box.  If  the  box  is  kept  from  sliding  by  a  small  obstacle 
placed  at  the  exterior  edge,  which  does  not  interfere  with  rota- 
tion, the  box  will  not  overturn  from  the  thrust  of  the  sand. 

By  the  Rankine  theory,  the  earth  thrust  against  the  vertical 
side  of  the  box  is  parallel  to  the  free  surface  or  horizontal.  This, 
combined  with  the  weight  of  the  box,  which  is  practically  negli- 
gible, gives  a  resultant  nearly  parallel  to  the  base,  so  that  over- 
turning should  occur.  Since  the  wall  is  stable,  the  evidence  is 
conclusive  that  the  assumed  direction  of  the  thrust  is  not  the 
true  one.  But  if  we  assume  the  friction  coefficient  of  sand  on 
wall  as  about  yz,  then  the  resultant  of  the  earth  thrust,  in- 
clined below  the  horizontal,  at  the  angle  whost  Jtangent  is  ^, 
with  the  weight  of  wall,  acting  along  the  vertical  through  its 
center  of  gravity,  will  pass  inside  the  base  of  the  wall  and  stabil- 
ity is  assured.  The  wall-friction  method  is  thus  not  inconsistent 
with  the  facts  if  the  friction  coefficient  is  about  ^.  It  was  not 
given  by  Flamant,  who  recorded  the  observation. 

The  author  has  investigated  the  results  of  experiments 
pertaining  to  the  case  of  retaining  walls  or  boards  at  the  limit 


13,14]  DIRECTION  OF  THRUST  19 

of  stability  and  has  recorded  his  conclusions  in  a  paper  en- 
titled "  Experiments  on  Retaining  Walls  and  Pressures  on 
Tunnels,"*  to  which  the  reader  is  referred  for  quantitative  re- 
sults. The  conclusion  is  drawn  from  the  experiments  cited 
on  retaining  walls  that  the  direction  of  the  thrust  as  given  by 
Rankine  is  never  experienced  in  walls  at  the  limit  of  stability, 
except  where  the  surface  slopes  at  the  angle  of  repose.  When, 
however,  the  wall-friction  is  included,  so  that  the  thrust  is  sup- 
posed to  make  the  angle  <p'  (when  $  ^  p)  or  <p  (when  <p  >  <p) 
with  the  normal  to  the  wall,  the  results  agree  fairly  well  with  the 
experiments,  the  backing  in  every  case  being  clean  sand.  In 
the  case  of  the  rotating  boards,  the  models  were  so  small  that 
cohesion  appreciably  affected  the  results.  But  with  an  assumed 
cohesion  of  i  Ib.  per  sq.  ft.,  it  was  found  that  the  direction  of 
the  thrust  was  approximately  as  given  by  the  last  method-t 

14.  Now,  if  it  is  granted  that  the  wall-friction  method  best 
explains  the  results  of  experiments  on  walls  at  the  limit  of  stability, 
it  is  still  doubted  by  some  whether  this  friction  can  always  be 
relied  on  in  the  case  of  a  stable  wall  throughout  the  years  of  its 
existence.  Let  us  consider  the  somewhat  analogous  case  of 
the  narrow  railroad  embankment  over  which  pass  heavy 
locomotives  at  great  speed.  The  very  integrity  of  this  embank- 
ment is  dependent  on  the  permanence  of  the  friction  coefficient, 
and  it  never  fails  when  the  foundation  is  good  and  it  is  well 
drained,  in  spite  of  the  vibration  to  which  it  is  subjected.  Or 
again,  take  the  wheat  in  a  tall  bin.  Hundreds  of  experiments 
have  shown  that  most  of  the  weight  of  the  wheat  is  carried  by 
friction  by  the  sides  of  the  bin  even  when  the  wheat  is  running 
out.  Again,  the  friction  factor  is  permanent  and  the  bin  would 
be  quickly  destroyed  if  it  was  not  so. 

But,  it  may  be  argued,  may  not  very  heavy  rains,  which  the 
drains  provided  may  not  carry  off  quickly  enough,  decrease 
<f>  and  tp',  and  thus  not  only  lead  to  an  increase  of  the  thrust  but 
likewise  cause  it  to  make  a  less  angle  with  the  horizontal  than 
before? 

*  Trans.  Am.  Soc.  C.  E.,  vol.  LXXII,  p.  403. 

t  See  Appendix  II  for  a  discussion  of  a  few  of  the  experiments. 


20  LAWS   OF   FRICTION   AND   COHESION 

Let  us  consider  the  two  possibilities  separately. 

From  a  consideration  of  the  angles  of  repose  in  Tables  I 
and  V,  it  is  seen,  for  an  average  clean  sand,  that  the  values  of  ??, 
for  the  states  dry,  moist,  wet  and  submerged,  will  be  something 
like  the  following: 

33°  4i',        44°,        33°  4i',         26°  34', 
and  for  ordinary  earth,  say, 

37°,        39°,        30°,         17°. 

For  hard  rock  or  rip-rap,  there  is  but  little  change.  For 
gravel  or  mixtures  of  sand,  clay,  and  gravel,  the  changes  are  not 
so  pronounced  as  in  the  case  of  ordinary  earth. 

From  these  results,  it  is  seen,  for  well-drained  filling,  that  for 
average  materials  there  is  an  appreciable  decrease  in  $  as  the 
material  changes  from  dry  to  wet.  This  could  be  met  by  de- 
signing the  wall  for  the  wet  filling,  in  the  first  instance.  Dock 
walls  should  of  course  be  designed  for  the  values  of  <p  correspond- 
ing to  filling  under  water,  for  the  part  of  the  filling  that  is  sub- 
merged and  for  wet  filling  for  the  part  above  that.  But  where  it 
is  only  in  times  of  excessive  floods  that  an  ordinary  retaining 
wall  is  submerged  or  partially  submerged,  it  seems  unnecessary  to 
design  such  a  wall  as  one  would  a  dock  wall.  The  failures  from 
such  excessive  floods  are  too  *few  to  warrant  the  extra  cost; 
in  fact,  it  will  be  more  economical  to  use  loose  rock  filling,  in 
part  at  least,  for  the  value  of  <p  for  such  filling  is  about  the. same, 
dry,  wet,  or  submerged. 

As  to  the  second  possibility,  it  seems  reasonable  to  suppose 
that  heavy  rains  may  lubricate  the  wall  to  such  an  extent  that 
(p  is  materially  decreased.  On  that  account  and  to  allow,  too, 
for  vibration,  due  to  heavy  passing  loads,  it  is  recommended  to 
multiply  only  the  normal  component  of  the  thrust  E\  by  the 
factor  of  safety  <r.  Then  on  combining  a  EI  with  the  friction 
component  EI  tan  p  (or  EI  tan  <p,  when  <pf  >  <?),  and  this  re- 
sultant in  turn,  with  the  weight  of  the  wall,  acting  along  the 
vertical  through  its  center  of  gravity,  let  the  wall  be  given 
such  a  thickness,  that  this  final  resultant  shall  pass  exactly 
through  the  outer  toe.  So  far  as  the  direction  of  the  thrust  is 


14, 15]  UNIT  STRESSES  21 

concerned,  this  method  is  equivalent  to  replacing  tan  <p'  by 
tan  <p'/<r.  Thus  if  a  —  3,  for  walls  over  10  ft.  high,  only  one- 
third  of  the  full  friction  is  included  in  the  moment  formula. 
This  seems  ample  in  allowing  for  heavy  rains  and  perhaps 
vibration. 

We  know  very  little  as  to  the  influence  of  vibration  on  earth 
thrust.  It  may  cause  minute  movements  of  the  wall  from  time 
to  time,  from  which  it  does  not  recover.  In  fact,  if  the  filling 
follows  up  such  movements,  it  is  difficult  to  see  how  the  wall  can 
regain  its  original  position. 

Trautwine*  says,  with  reference  to  his  little  wooden  retaining  walls,  about 
6  inches  in  height  and  backed  by  sand:  "We  found  that  the  tremors  produced 
T}y  passing  vehicles  in  the  street,  by  the  shutting  of  doors  and  walking  about 
the  room,  sufficed  to  gradually  produce  leaning  in  walls  of  considerably  more 
than  twice  the  mere  balancing  stability  while  quiet."  The  models  were  in 
the  upper  room  of  a  strongly  built  house. 

This  observed  increased  leaning  with  time  of  actual  retaining 
walls  may  be  largely  due  to  the  vibration  caused  by  heavy 
passing  loads  and  certainly  the  factor  of  safety  for  walls  6  to 
10  ft.  high  should  be  larger  than  for  walls  over  10  feet;  say,  at 
least  3.5. 

It  is  a  common  and  commendable  practice  to  step  the  back 
of  a  retaining  wall,  as  illustrated  in  Fig.  6.  Any  moving  over  of 
the  top  of  the  wall  will  tend  to  cause  a  break  in  the  earth  along 
a  plane  touching  the  back  of  the  wall,  since  the  triangular  por- 
tions of  earth  over  the  steps  will  move  with  the  wall.  The 
friction  developed  is  thus  of  earth  on  earth,  so  that  <?'  is  re- 
placed by  <p.  *If  <p  for  a  smooth  concrete  wall,  lubricated  with 
water  in  times  of  heavy  rains,  is  much  smaller  than  <p  for  the 
wet  earth,  there  is  decided  economy  in  stepping  the  wall,  since 
the  direction  of  the  thrust  on  the  back  face,  is  more  inclined 
to  the  horizontal. 

15.  Unit  Stresses  on  a  Horizontal  Section.  Let  A  D  E  B, 
Fig.  6,  represent  a  portion  of  a  retaining  wall  contained  between 
two  vertical  parallel  planes,  one  foot  apart,  and  perpendicular 
to  the  front  and  rear  faces.  In  fact,  throughout  the  book,  one 

*  Engineers'  Pocket  Book,  igth  ed.,  p.  605. 


22 


LAWS  OF  FRICTION  AND  COHESION 


foot  length  of  wall  will  always  be  understood,  when  the  cross- 
section  is  uniform,  or  there  are  no  buttresses  or  counterforts. 

By  methods  to  be  given  later,  the  earth  thrust  on  the  back 
face  AD  can  be  estimated.  Both  the  Rankine  and  the  wall- 
friction  methods  will  be  given,  so  that  the  reader  can  take  his 
choice — suppose  the  latter  method  to 
be  employed;  then  if  the  rear  face  is 
plane,  compute  the  thrust  on  it  for  the 
kind  of  earth  supposed,  the  thrust  ma- 
king the  angle  <p'  (when  <p  ^L  <p)  or  <p 
(when  $  >  <p)  with  the  normal  to  the 
rear  face;  but  if  the  rear  face  is  stepped, 
compute  the  thrust  on  a  plane,  say  AD, 
having  about  the  average  batter  of  the 
face,  the  thrust  making  the  angle  (p  (of 
earth  on  earth),  with  the  normal  to  AD. 
The  earth  thrust,  acting  on  the  rear 
face,  at  one- third  the  height  of  wall,  is 
produced  to  meet  the  vertical  through 
the  center  of  gravity  of  the  wall  at  G. 
On  laying  off  G  H,  to  scale,  to  represent 
the  weight  of  the  wall  and  G  M  to  repre- 
sent the  earth  thrust,  and  drawing  MN 
\\GH,HN\\  GM,  to  intersection  N,  the 
diagonal  NG  of  the  parallelogram  is  seen  to  represent  the  re- 
sultant on  the  horizontal  section  A  B.  Let  us  suppose  that  it 
cuts  this  section  a»t  /,  a  distance  a  feet  to  the  left  of  its  center  C. 
Also  let 

F  =  vertical  component  of  the  resultant  G  Yf  on  A  B, 

I  =  length  of  A.  B  in  feet. 

If  we  decompose  at  the  point  /  the  resultant  NG  into  a 
horizontal  component  and  a  vertical  component  F,  the  former 
is  resisted  by  the  shear  along  the  section  AB  of  the  wall,  or  if 
AB  is  the  bottom  of  the  wall,  it  is  resisted  by  the  friction  between 
the  wall  and  the  earth  on  which  it  rests.  There  remains  the 
vertical  component  F  to  consider.  At  the  center  C  of  the 
section  AB,  conceive  applied  two  vertical  opposed  forces,  each 


FIG.  6 


15]  UNIT  STRESSES  23 

equal  to  F.  This  does  not  affect  equilibrium.  The  force  F 
acting  downward  at  C  may  be  supposed  to  be  the  resultant  of  a 
uniformly  distributed  stress,  pi  =  F/l,  on  the  section,  as  shown 
by  the  arrows  just  below  AB.  The  force  F  at  /,  acting  down, 
with  the  remaining  force  F  at  C,  acting  up,  constitutes  a  couple 
whose  moment  is,  F.  1C  =  Fa. 

For  a  monolithic  wall,  with  front  and  rear  faces  vertical, 
this  couple  will  cause  a  uniformly  varying  stress  (as  shown  by 
the  little  arrows  just  below  the  first)  whose  intensity  at  A  and      •> 
B  is,  by  known  laws, 

Fa       6  Fa 


Evidently  when  /  is  to  the  left  of  C,  pz  is  compressive  at  B  and 
tensile  at  A  .  The  total  unit  stress  at  B  and  A  is  thus, 

F  (        6a\ 

P  =  pi  *=  #2  =  y  (i  *  -j)  .     .,     .     .     .      (3) 

the  +  sign  referring  to  the  point  B,  the  —  sign  to  the  point  A. 
The  lower  arrows,  contained  in  a  trapezoid,  represent  the  dis- 
tribution of  the  stress,  corresponding  to  what  is  called  the 
"trapezoid  law,"  F  is  the  resultant  in  position  and  magnitude 
of  these  stresses.  The  same  formula  (3)  is  supposed  to  hold 
when  AB  is  the  plane  of  contact  of  the  wall  with  the  earth,  the 
values  of  p  giving  the  vertical  components  of  the  "soil  reactions'7 
at  B  and  A  (acting  of  course  upward)  when  the  entire  base  is 
in  compression.  It  will  be  observed  from  (3),  that  p  is  positive 
at  A  when  a  <  1/6  or  when  the  resultant  on  AB  cuts  it  within 
its  "middle  third." 

F         A 

When  a  =  1/6,  p  =  o  at  A   and  p  =  2  7  at  A,.     Thus  the 

unit  pressure  at  A  is  double  the  average  and  the  trapezoid 
reduces  to  a  triangle. 

When  a  >  1/6,  or  the  resultant  on  AB  passes  outside  of  its 
middle  third,  then,  when  the  masonry  can  resist  the  tension, 
the  stress  at  A  is  minus,  indicating  tension  and  is  given  by  (3). 
Since  the  soil  can  only  resist  compression,  formula  (3)  does  not 


24 


LAWS   OF  FRICTION  AND   COHESION 


apply  to  finding  soil  reaction  when  a  >  1/6.  For  this  case 
and  also  for  a  masonry  section  which  cannot  resist  tension,  the 
unit  stresses  are  supposed  to  vary  as  the  ordinates  of  a  triangle. 
Hence,  since  F  is  their  resultant,  it  must  act  through  the  center 
of  gravity  of  the  triangle  and  the  section  will  sustain  compres- 
sion only  over  the  length,  ^.BI,  Fig.  7,  and  the  maximum  in- 
tensity at  B  is  double  the  mean,  or, 


(4) 


on  putting  BI  =  b.     The  stress  on  AB,  at  the  distance  3  BI  = 


no 


FIG.  7 

3  b,  from  B  is  zero,  and  to  the  right  of  this  point  there  is 
stress  and  the  joint  there  will  open. 

The  formulas  above  have  been  derived  for  rectangular  walls. 
They  are  approximately  true  for  battered  walls,  where,  as  usual, 
the  batter  does  not  exceed  a  few  inches  per  foot,  and  they  will  be 
used  for  the  ordinary  types  of  walls  throughout.  The  value 
given  above  for  p2  is  strictly  true  only  for  rectangular  walls  and 
departs  more  and  more  from  the  truth  as  the  batter  increases, 


15,  16]  FACTORS   OF   SAFETY  25 

so  that  the  trapezoid  law  is  only  approximately  true.  From 
the  experiments  on  india-rubber  dams,  mentioned  in  Appen- 
dix I,  it  is  seen  that  the  vertical  components  of  the  stresses  at  the 
outer  toes  are  less  than  the  values  corresponding  to  the  trapezoid 
law,  so  that  the  approximation  is  on  the  side  of  safety.  In 
reality,  the  stresses  are  greater  in  the  body  of  a  retaining  wall 
(or  dam),  and  less  at  the  faces,  than  the  theory  calls  for. 

16.  Factors  of  Safety  Against  Overturning  and  Sliding. 
The  factor  of  safety  against  overturning  is  generally  defined  to 
be  the  number  by  which  the  earth  thrust  GM,  Figs.  6  and  7, 
must  be  multiplied,  so  that  the  increased  thrust,  when  com- 
bined with  the  weight  of  the  wall  GH,  shall  give  a  resultant 
•that  passes  through  B,  the  outer  toe  of  the  wall.  See  Art.  14 
for  another  definition,  which  applies  more  particularly  when 
the  wall  friction  is  included  in  computing  the  thrust. 

To  define  the  factor  of  safety  against  sliding,  consider  Fig.  7, 
where  GM  represents  the  earth  thrust,  GH  the  weight  of  wall, 
and  GU  =  F  the  vertical  component  of  the  resultant  GN  on 
the  base  AB. 

Lay  off  the  angle  GrV  with  rV,  the  normal  .to  AB,  equal  to 
<?",  the  angle  of  friction  of  masonry  on  earth.-  Extend  Gr  to 
meet  HN  produced  at  P  and  draw  PL  vertical  to  meet  GM 
produced  at  L  and  PT  horizontal  to  meet  GH  produced  at  T. 
Then  if  the  earth  thrust  was  GL,  the  resultant  on  the  base 
would  be  GP  and  sliding  would  be  impending.  Hence  the 
factor  of  safety  s  against  sliding  is, 

GL       PT 

"  G&~  NU' 

If  we  extend  UN  to  R  to  meet  GP,  we  note  that  the  total  friction 
that  can  be  exerted  along  the  base  is,  F  tan  <?"  =  GU  tan  <p"  = 
UR.  The  ratio  of  this  to  the  horizontal  component  of  the  thrust 
NU  is  sometimes  given  as  the  factor  of  safety;  call  it  /. 

GU  tan  y"         tan  <'  tan  <?' 

NU         ~  NU/GU  ~~  tanGNM  ' 

We  always  have  /  <  s,  except  when  the  thrust  is  horizontal, 


26  LAWS   OF  FRICTION  AND  COHESION 

when  they  are  equal.  The  last  formula  supposes  the  vertical 
component  of  the  earth  thrust  to  remain  constant  when  the 
horizontal  component  NU  is  multiplied  by  /  to  give  RU  = 
F  tan  <p".  The  formula  thus  gives  a  slight  excess,  and  it  is 
often  the  most  convenient  formula  to  use.  If  practicable,  the 
factor  of  safety  against  sliding  should  be  2.  As  this  is  often 
difficult  to  attain,  recourse  is  had  to  projections  of  masonry  on 
the  base;  otherwise  the  base  is  tilted  up,  so  that  GN  makes  a 
sufficiently  small  angle  with  the  normal  to  the  base,  always 
much  less  than  <p". 

17.  Additional  Remarks  Relative  to  the  Middle  Third 
Requirement  and  the  Use  of  Eq.  (3).  In  Art.  9,  Table  IV, 
the  bearing  capacity  of  various  soils  is  given.  If,  for  a  partic- 
ular soil,  the  maximum  pressure  found  from  eq.  (3)  is  greater 
than  is  permissible,  then  the  base  must  be  made  larger.  This  is 
generally  effected  by  building  a  footing  or  foundation  course 
projecting  beyond  the  front  face  of  the  wall.  In  this  way,  too, 
the  resultant  on  the  earthen  base  can  be  brought  very  near  its 
center.  To  cause  this  resultant  to  pass  near  the  center  may 
entail  an  extra  large  footing,  but  if  the  additional  expense  is 
not  prohibitive,  it  should  positively  prevent  that  increased  lean- 
ing with  time,  observed  in  the  case  of  some  walls. 

It  is  usually  specified  that  the  resultant  on  the  base  of  a 
wall  shall  cut  it  within  its  middle  third.  In  case  of  a  limited 
right  of  way,  it  may  be  allowed  to  pass  slightly  outside  of  this 
limit  in  the  case  of  a  rock  foundation,  since  stability  is  assured. 
But  for  ordinary  earth,  particularly  if  the  soil  resistance  is 
uncertain,  the  requirement  should  generally  be  observed,  and  it 
is  further  advised  to  build  the  projecting  footing  course  men- 
tioned, so  that  the  resultant  on  the  base  shall  ffass  as  near  its 
center  as  is  practicable,  thus  ensuring  a  more  nearly  uniform 
pressure  on  the  base. 


CHAPTER  II 

NON-COHESIVE  EARTH.      GRAPHICAL  METHODS 

18.  Surface  of  Rupture.  A  large  number  of  interesting 
experiments  were  made  by  Leygue*  on  boards  backed  by  clean 
sand,  having  but  little  cohesion,  to  determine  the  surface  of 
rupture  in  the  sand  when  the  retaining  boards  were  rotated 
about  their  lower  edges.  The  box  in  which  the  sand  was  con- 
ined  had  glass  sides,  the  pivoted  boards  in  front  representing 
the  retaining  walls.  To  observe  the  relative  movement  of  the 
grains  of  sand  as  the  wall  moves  over,  thin  horizontal  strata 
of  pulverized  white  plaster  were  placed  in  the  sand  at  various 
leights.  As  the  retaining  board  moved  over,  it  was  observed 
that  all  the  sand  particles  lying  between  the  retaining  board 
and  a  surface  convex  upward  and  passing  through  the  foot  of 
the  board,  moved  parallel  to  this  surface.  This  surface,  called 
he  surface  of  rupture,  remained  invariable  as  the  rotation  was 
ncreased  up  to  a  certain  point  (30°  for  a  horizontal  surface  of 
earth) ;  but  on  completing  the  rotation  by  lowering  the  retaining 
)oard  to  the  horizontal,  the  surface  moved  to  the  rear  and  the 
earth  finally  took  its  natural  slope. 

The  same  results  were  found  to  hold  when  the  retaining 
)oard  was  moved  parallel  to  itself,  except  that  the  surface  of 
rupture  tended  from  the  start  to  move  to  the  rear. 

It  was  further  observed  that  the  full  friction  of  the  earth 
was  exerted  against  the  wall  in  either  movement. 

The  height  of  the  retaining  boards  was  varied  from  0.66  to 
.82  feet  for  the  first  series  of  experiments,  the  boards  being  first 
put  at  the  inclination  desired  before  rotating. 

In  the  next  set  of  experiments  the  boards  were  3.28  and  6.56  ft. 
high,  with  the  filling  level  with  the  top  and  also  with  a  surcharge 


*  Annales  des  Fonts  et  Chaussees,  Nov.,  1885. 
27 


28  NON-COHESIVE   EARTH.      GRAPHICAL   METHODS 

up  to  3.28  feet.  It  can  be  shown  that  the  influence  of  cohesion  is 
very  small  for  the  sand  or  broken  stone  used  for  the  walls  6.56 
ft.  high.  For  the  larger  heights,  the  surface  of  rupture  was  more 
irregular  than  for  the  smaller  heights,  but  it  was  always,  taking 
the  average,  convex  upward  (or  concave  downward)  not  only 
for  the  sand,  but  also  for  the  stones  which  varied  from  1.5  to  20 
inches  in  diameter.  The  convexity  of  the  surface  of  rupture 
was  only  slight,  but  it  was  invariable. 

For  consolidated  earth,  it  is  well  known  from  observation  of 
the  caving  of  banks  that  the  surface  of  rupture  is  concave  up- 
ward. Such  earth  is  largely  endowed  with  cohesion. 

In  this  chapter,  cohesion  will  be  neglected  and  the  surface  of 
rupture  will  be  taken  as  plane.  The  theory  developed,  for  the 
cases  represented  by  the  experiments  at  least,  will  thus  be 
approximate,  but  it  is  probably  a  close  approximation  for  clean 
sand  or  rock  filling,  for  walls  6  to  10  ft.  high  and  upward.  The 
cohesion  is  negligible  for  such  heights,  and  the  surface  of  rupture, 
for  such  materials,  is  nearly  plane. 

For  fresh  earth  deposited  behind  a  retaining  wall,  there  is 
much  more  cohesion  than  for  clean  sand,  Art.  7,  so  that  the 
computed  thrust,  neglecting  cohesion,  will  be  somewhat  in 
excess,  particularly  for  heights  of  wall  of  only  6  to  10  ft.,  and 
this  excess  will  become  larger  with  time  (except  for  the  influence 
of  heavy  rains)  since  the  cohesion  generally  increases  with 
time. 

19.  Earth  Endowed  with  Friction  but  Devoid  of  Cohesion. 
Sliding  Wedge  Theory.  Let  AB,  in  Fig.  8,  represent  the  inner 
face  of  a  retaining  wall,  backed  by  earth  whose  surface  is 
BCD. 

The  angle  of  friction  of  the  earth  =  <p,  the  angle  of  friction 
of  earth  on  wall  =  <?'.  Let  weight  of  earth  per  cubic  foot  =  w. 
As  usual,  one  foot  length  of  wall  perpendicular  to  the  plane 
of  the  paper  will  be  considered.  Suppose  the  earth  tends  to 
slide  down  a  plane  of  rupture  AC.  The  weight  of  the  wedge 
of  rupture  ABC  =  w  X  area  ABC  =  W.  As  shown  in  Art.  12, 
the  reaction  E  of  the  wall  (equal  and  opposed  to  the  earth 
thrust  on  the  wall)  makes  the  angle  <p  with  the  normal  to  AB 


9,20] 


ACTIVE    AND    PASSIVE    THRUST 


29 


hen  </  <_  <p  or   the  angle   <?  when   $   >  </>.     See  Art.  30  for 
xceptions  to  this  rule. 


FIG.  8 

In  the  case  of  experimental  walls,  $  should  be  determined 
y  experiment.  For  an.  actual  rough  masonry  wall,  where  ^' 
>  not  known,  it  is  usually  replaced  by  (p.  For  stepped  walls, 
'  is  always  to  be  replaced  by  <p. 

To  allow  for  the  lubrication  of  the  wall  with  water  or  for 
xcessive  vibration,  some  designers  take  $  =  o.  See  Arts.  12-14. 
'or  walls  leaning  toward  the  earth,  the  direction  of  the  thrust 
an  be  taken  approximately  as  above  when  the  inclination  of 
he  wall  to  the  vertical  does  not  exceed  about  10°.  For  greater 
iclinations  see  Art.  62. 

The  reaction  of  the  earth  on  AC  consists  of  a  normal  com- 
onent  N  and  a  frictional  component  ^V  tan  <p,  acting  up  along 
1C,  the  resultant  of  the  two^  thus  making  an  angle  <p  with 
tie  normal  to  AC.  To  find  "easily  the  direction  of  this  re- 
ultant  reaction,  lay  off  HAD  =  <p  and  let  DAC  =  ft.  Then, 
i  from  any  point  g  in  a  vertical  through  A,  we  lay  off  Ags  =  ft 
t  will  follow  that  gs  is  the  direction  of  the  resultant  reaction 
n  AC.  This  follows,  because  if  gl  is  perpendicular  to  AC  .'. 
Igl  =  HAC,  the  sides  being  perpendicular  each  to  each.  Hence 
[gl  =  ft  +  <f>,  and  if  we  lay  off  Ags  =  ft,  then  Igs  =  <f>  or  gs 
3  inclined  at  the  angle  <p  with  the  normal  to  AC. 

This  reaction  is  not  drawn  in  true  position,  since  it  must 
>ass  through  the  intersection  of  E  and  W,  the  three  forces, 
2,  W,  and  the  reaction,  being  in  equilibrium. 

20.  Active  and  Passive  Thrust.  The  thrust  £,  Fig.  8, 
aused  by  the  tendency  of  the  mass  ABC,  from  its  weight,  to 


30 


NON-COHESIVE  EARTH.       GRAPHICAL   METHODS 


slide  down  the  plane  AC,  is  called  the  active  thrust  of  the  earth 
against  the  wall.  Here  the  friction  component  of  the  reaction 
of  AC  acts  up  or,  as  usual,  always  opposed  to  the  impending 
motion. 

If,  however,  the  wall  is  subjected  to  a  force  acting  toward 
the  earth,  which  exceeds  the  active  thrust,  it  will  bring  into 
play  the  passive  resistance  of  the  earth  to  sliding  up  some 
plane,  generally  below  the  plane  of  rupture  for  active  thrust. 

When  the  force  applied  to  the  wall  is  so  great  that  motion 
up  some  plane  as  AC  is  impending,  then  the  frictional  com- 
ponent of  the  reaction  of  AC,  N  tan  <p,  acting  opposed  to  the 
impending  motion,  will  act  down,  so  that  the  resultant  reaction 
now  lies  above  the  normal  to  AC  and  makes  the  angle  <p  with  it. 


a  2 


FIG.  9 


The  force  acting  on  the  wall  is  equal  and  directly  opposed 
to  the  resistance  or  so-called  passive  thrust  of  the  earth  on  the 
wall. 

21.  Graphical  Determination  of  Active  Earth  Thrust 
Against  a  Wall.  In  Fig.  9,  let  AB  represent  the  inner  face 
of  a  wall,  backed  by  earth  whose  surface  Bb8  has  a  uniform 
slope.  Let  us  suppose  <?'  <_  #  and  that  AB  lies  at  or  above 


21]  ACTIVE  EARTH   THRUST  31 

the  " limiting  plane"  of  Art.  27.  Let  the  vertical  Ag  intersect 
the  free  surface  at  bi  and  lay  off  bj)2  =  bjbZl  etc.,  along  bj)%* 
Draw  the  lines  Ab2,  Abs,  etc.,  to  represent  possible  planes  of 
rupture.  Let  the  length  of  the  perpendicular  from  A  upon 
\Bbs  =  p\  then  the  weight  of  the  prism  ABbi  =  %  w.p.  Bbi, 
weight  of  AbJ)2  =  weight  of  Ab2bs  =  etc.  =  %  w.p.  bj)2.  Then, 
by  addition,  we  can  compute  the  weights  in  pounds  of  the 
[successive  trial  prisms  of  rupture  ABb2,  ABb3)  etc. 

On  any  convenient  vertical  as  Ag,  lay  off  ggi  =  weight  of 
ABbi,  gg2  =  weight  of  ABb2,  ggs  =  weight  of  ABb3j  etc. 

With  A  and  g  as  centers  and  any  convenient  common  radius 
describe  arcs  Ha^  and  As2.  Lay  off  the  angle  HAd  =  <p  and  with 
dividers  lay  off  chords  As2j  As3,  As*,  etc.,  equal  to  chords  da^, 
,z,  da>4,  etc.  The  points  #2,  #3,  etc.,  are  where  the  lines,  Ab2) 
bs,  etc.,  produced  when  necessary,  cut  the  arc  Ha2. 

Since  the  construction  gives  Z  Ags2  =  Z  dAck,  etc.,  by 
Art.  19,  it  is  seen  that  gs2,  gSz,  etc.,  make  the  angle  <p  with  the 
normals  to  the  planes  Ab2,  Abs,  etc.,  respectively.  Hence  gs2, 
S)  etc.,  have  the  directions  of  the  reactions  of  the  successive 
trial  planes  of  rupture  Ab2,  Ab3,  etc. 

Next  draw  a  line  inclined  at  the  angle  <pf  to  the  normal  to 
AB  to  represent  the  direction  of  E  as  shown  and  through  gi, 
gz,  gz}  etc.,  draw  lines  parallel  to  E.  Let  such  lines  through 
,  gs,  . . . ,  gs  intersect  gs2,  gsB,  .  .  . ,  gs8,  at  c2,  cs,  .  .  . ,  cs. 

Now  observe,  for  any  supposed  plane  of  rupture  Ab3,  that 
the  corresponding  prism  of  rupture  ABb3A  is  held  in  equilibrium 
by  its  weight,  the  reaction  of  the  wall  and  the  reaction  of  the 
plane  Ab3  and  that  these  three  forces  are  represented  to  scale 
by  the  sides  gg3,  g3cs  and  csg  of  the  force  triangle  gg3c3.  Similarly 
for  a  supposed  plane  of  rupture  Ab±.  the  forces  acting  on  the 
corresponding  wedge  of  rupture  ABb±A,  are  given,  to  scale,  by 
the  sides  of  the  triangle  gg^c^  gg±  representing  its  weight,  gtd 
the  reaction  E  of  the  wall  and  c$g  the  reaction  of  the  plane 

*  It  is  usually  best  to  lay  off  from  B  along  Bb$  equal  distances  of  I  or  2 
feet,  so  that  the  weights  of  the  prisms  ABbi,  Abib2,  etc.,  are  all  equal,  and  the 
weights  of  the  successive  prisms,  ABb\,  ABb2,  etc.,  can  be  found  by  simple 
addition.  In  the  figure,  Ab\  was  taken  as  vertical  simply  for  purposes  of 
comparison  with  the  results  pertaining  to  Fig.  10. 


32  NON-COHESIVE   EARTH.      GRAPHICAL  METHODS 

Ab±,  since  gc4s4  was  drawn  making  the  angle  <p  with  the  normal 
to  the  plane  Ab±  and  is  thus  parallel  to  its  reaction.  Of  all  the 
trial  prisms  of  rupture,  there  can  be  only  one  true  one  with  its 
resulting  thrust,  and  this  corresponds  to  the  greatest  of  the 
lines  #2^2,  gtfz,  g&±,  etc.,  which,  in  this  example,  is  g4c4j  and  thus 
represents  the  true  value  of  E. 

This  follows  from  Rankine's  principle  of  Art.  5,  because  for 
any  less  thrust  than  g4c4,  the  point  c4  will  fall  to  the  left  of  its 
present  position,  hence  the  new  gc4  will  make  an  angle  greater 
than  <p  with  the  normal  to  the  plane  Ab±,  which  is  inconsistent 
with  the  laws  of  stability  of  a  granular  mass  as  stated  in  Art.  5. 

On  drawing  a  curve  through  c2,  c3,  .  .  .  ,  c8,  and  a  vertical  tan- 
gent /i/s  to  it,  the  greatest  of  the  lines  g2c2,  gsCs,  .  .  .  ,  can  be 
readily  found.  In  any  practical  example,  it  is  well  to  draw 
other  trial  planes  of  rupture  near  Ab±,  the  one  corresponding 
to  g4c4,  to  determine  by  the  construction  more  accurately  the 
true  thrust  and  the  true  plane  of  rupture.  In  fact,  after  making 
a  number  of  constructions  similar  to  the  above,  for  various 
inclinations  of  AB  and  Bb8,  the  probable  position  of  the  true 
plane  of  rupture  can  be  guessed  at  rather  closely,  so  that  only 
a  few  trial  planes  need  be  drawn  near  this  position.  It  is  pos- 
sible that  the  plane  of  rupture  determined  by  this  construction, 
may  differ  in  direction  i°  or  2°  from  the  true  one,  but  the  earth 
thrust  is  determined  more  accurately  than  by  any  other  graphical 
method,  since  intersections  c2,  c3,  .  .  . ,  are  determined  very  closely, 
the  angles  g2Czg,  g3csg,  .  .  .  ,  being  large.*  The  principle  demon- 
strated above  that  the  true  active  earth  thrust  against  the  wall 
is  the  maximum  of  the  thrusts  corresponding  to  the  trial  prisms 
of  rupture  was  originally  given  by  Coulomb  as  a  self-evident  first 
principle  and  is  known  as  Coulomb's  "wedge  of  maximum 
thrust." 

Having  found  the  true  reaction  E  of  the  wall  (equal  and 
opposed  to  the  earth  thrust  on  the  wall)  which  equals  g4c4  to 
scale  and  the  plane  of  rupture  Ab±,  to  find  the  true  resultant 

*  The  method  used  above  for  ascertaining  the  trial  thrusts  is  analogous  to 
that  first  given  by  Prof.  H.  T.  Eddy  in  "New  Constructions  in  Graphical 
Statics." 


21-23]  ACTIVE  EARTH   THRUST  33 

thrusts  on  the  planes  A bi,  Ab2,  .  .  . ,  Abs,  we  extend  the  lines 
through  gi,  gz,  . .  .  ,  g8)  that  are  parallel  to  E,  to  /i/8,  meeting  it 
at  the  points  /i,  fe,  .  .  .  ,  t8.  Then  gfa  =  g^  =  #&  =  £4^4,  etc.; 
hence  since  the  resultant  of  ggi  and  gi/i  gives  the  resultant  thrust 
on  the  plane  A  bi  and  similarly  for  the  other  planes,  it  follows, 
if  the  straight  lines  tig,  kg,  .  .  .  ,  kg,  are  drawn,  they  represent  the 
actual  reactions  on  the  planes  Abi,  Ab2,  ...  ,  Abs,  respectively. 

These  lines  (not  drawn  in  the  figure),  other  than  dg}  all  make 
less  angles  with  the  horizontal  than  the  trial  values  of  the  reac- 
tions and  thus  make  less  angles  than  <p  with  the  normals  to  their 
corresponding  planes.  Hence  the  conditions  of  stability  are 
all  satisfied,  and  if  the  wall  gives,  sliding  will  occur  only  down 
the  plane  of  rupture  Ab±. 

To  ensure  accuracy  in  this  construction,  the  scales  of  force 
and  length  should  be  large  and  the  radius  AH  should  be  pref- 
erably 5  or  10  inches  or  more.  The  length  g4c4,  in  this  case, 
to  the  scale  of  force  gives  the  thrust  E  in  pounds — similarly  the 
length  of  tig,  to  the  scale  of  force,  gives  the  thrust  on  the  vertical 
plane  Ab,  in  pounds.  It  is  seen  not  to  be  parallel  to  Bb& ;"hence 
it  does  not  agree  in  direction  with  the  Rankine  thrust  on  a 
vertical  plane,  which  is  parallel  to  the  top  slope. 

22.  Variation  of  E  with  <?'.     In  Fig  9,  if  we  suppose  <p'  to 
diminish,  the  lines  #2^2,  g&z,  etc.,  parallel  to  E,  will  approach 
nearer  the  horizontal,  the  points  c  will  recede  from  Ag,  hence 
the  thrust,  represented  by  the  greatest  of  the  lengths  g2c2,  gsC3, 
.  .  .  ,  for  a  given  <?',  will  have  a  greater  horizontal  component 
than  before.     On  the  contrary,  if  </  increases,  the  points  c  will 
approach  the  vertical  Ag  and  the  horizontal  component  of  the 
thrust  will  diminish. 

23.  Examples.     In  all  the  examples  below,  take  <p  =  <?'  =  33°  41',  cor- 
responding to  a  slope  of  I  }4  base  to  I  rise. 

Ex.  i.  Let  the  inner  face  AB  of  a  wall  20  ft.  high  have  a  batter  of  3.75 
in.  to  the  foot  or  a  horizontal  projection  of  6.25  ft.  and  suppose  the  earth, 
weighing  w  =  100  Ib.  per  cu.  ft.,  to  have  a  free  plane  surface,  extending  from 
B  above  the  horizontal  and  making  an  angle  with  it  of  6°.  Prove  by  the  con- 
struction of  Fig.  9,  that  the  earth  thrust  on  AB,  inclined  at  the  angle  <?  to  its 
normal,  is  9560  pounds,  the  plane  of  rupture  meeting  the  surface  about  20  ft. 
from  B. 

Exs.  2.  In  the  table  are  given  the  values  of  E  for  six  different  batters  of 


34 


NON-COHESIVE   EARTH.      GRAPHICAL  METHODS 


AB,  for  <f>  =  <f>'  =  33°  41 ',  w  =  i  and  the  earth  surface  level  with  B,  the  vertical 
height  of  AB  being  10  ft. 


Batter  of  AB 

E 
(fora; 

=  1) 

K 

Ki 

o.  .  . 

j-i 

o 

o 

I'U) 

o  109 

I  in. 

to  T 

ft.  . 

JC 

I 

o 

ISI 

o  126 

2  in. 
3  in. 

to  I 
to  T 

ft  
ft  

17 

iq 

I 

7 

0 

o 

.171 

IQ7 

0.143 

o.  164 

4.  in. 

to  I 

ft 

22 

o 

224 

o  187 

5  in. 

to  i 

ft 

2C. 

2 

o 

252 

O   2IO 

To  find  the  value  of  E  when  the  earth  weighs  w  Ib.  per  cu.  ft.,  multiply  the 
tabular  value  by  w  to  get  the  thrust  in  pounds.  The  columns  K  and  Ki 
are  inserted  for  future  reference,  K  being  the  thrust  for  w  =  i  and  height  of 
AB  =  i,  KI  =  K  cos  <p,  being  its  component  normal  to  AB,  as  will  be  ex- 
plained in  Art.  24. 

Exs.  3.  In  the  next  table,  the  values  E  refer  to  walls  10  ft.  high,  whose 
tops  lean  toward  the  earth  the  amount  given,  expressed  as  in  the  case  of  batter. 
As  before,  the  earth  thrust  on  AB  is  supposed  to  make  the  angle  <f>  with  its 
normal;  (f>  =  <p'  =  33°  41',  w  =  i  and  the  free  surface  is  level  with  B. 


Lean  of  AB 

E 

(forw  =  i) 

K 

Ki 

i  in.  to 

I  ft.  . 

II.4 

o.  114 

0.095 

2  in  to 

I  ft 

0  7 

O  OQ7 

O.oSl 

The  values  of  K  and  K\  for  any  intermediate  batters  can  be  interpolated 
from  the  tables  with  sufficient  accuracy. 

If  the  student  will  do  careful  drawing,  he  should  be  able  to  reach  the  above 
results  on  a  drawing  that  can  be  contained  on  a  sheet  of  writing  paper.  When 
the  earth  slopes  downward  from  B,  the  construction  of  Fig.  9  equally  applies 
for  the  various  inclinations  of  AB. 

The  construction  of  Fig.  9  fails  for  cases  where  the  free  surface  approaches 
the  natural  slope,  since  the  plane  of  rupture  then  approaches  the  natural  slope. 
In  such  cases  we  have  to  resort  to  formulas  to  be  derived  la'^f.  ( 

24.  Center  of  Pressure.  In  Fig.  9,  let  h  =  vertical 
height  of  AB,  the  free  surface  of  earth  being  either  level  or 
having  a  uniform  slope,  ascending  or  descending  from  B.  Then 
for  the  same  inclination  of  AB,  as  h  varies,  the  plane  of  rupture 
remains  at  the  same  inclination,  hence  the  sections  of  the  prisms 
of  rupture  for  different  values  of  h  are  similar  triangles,  so  that 
the  thrusts  E,  which  vary  directly  with  the  weights  of  the 


24-26]  CENTER   OF   PRESSURE  35 

corresponding  prisms,  will  also  vary  as  the  areas  of  these  triangles 
or  as  the  squares  of  AB  or  as  h2.  Hence  the  thrust  can  be 
represented  by  an  expression  of  the  form, 

E  =  K  w  h2 

where  K  is  a  constant  for  a  particular  surface-  slope  and  inclina- 
tion of  AB.  It  follows,  as  in  the  case  of  water  pressure,  that 
the  resultant  E  is  applied  at  one-third  the  height  of  the  wall  from 
the  base  or  that  the  center  of  pressure  is  found  at  a  point  on 
AB,  (1/3)  ^^from^. 

25.  Definition  and  Use  of  Quantities  K  and  K\.     Referring 
to    the    above    formula,    we    see,    when    w  =  i,    h  =  i,    that 
E  =  K,  or  K  is  the  thrust  on  a  wall  i  //.  high,  when  w  =  i,  and 
that  its  normal  component  is  K  cos  <p,  or,  K  cos  <p  when  <pr  <  <p. 
Putting  KI  for  this  normal  component,  we  have, 

KI  =  K  cos  <PJ  or,  KI  =  K  cos  <p  (pf  <  <p). 

It  is  easy  to  find  K  by  the  previous  construction,  assuming 
w  =  i,  h  =  i;  or  we  may  assume  w  =  i,  h  =  io,  giving  the 
values  of  E  in  the  table  of  the  preceding  article.  Then  E  = 
Kh2  =  io2  K;  hence,  K  =  E/ioo..  In  this  way  the  values  of  K 
in  the  table  were  derived. 

From  this  table,  we  can  find  the  value  of  the  thrust  for  any 
height,  for  <p  =  <?'  =  33°  41',  earth  surface  level  from  B  and 
a  given  w  and  inclination  of  AB.  Thus  for  w  —  ioo,  a  batter 
of  3.75  inches  to  the  foot  and  h  =  io  ft., 

K  =  0.197  +  ZA  (-027)  =  0.217 

.'.  E  =  K  w  h*  =  0.217  X  ioo  X  io2  =  2170  Ibs. 
Similarly  for  the  same  batter  and  value  of  w,  but  for  a  height 
of  h  =  20  ft.,  as  in  Ex.  i,  Art.  23, 

E  =  0.217  X  ioo  X  2o2  =  8680  Ibs. 

26.  Thrust    on    a    Vertical   Plane    in    an   Unlimited  Mass 
of  Earth.     In  Fig.   io,  let  5  Obb  represent  the  free   surface- 
supposed   plane — and   let  AO  represent  a  vertical  plane  in  it, 
the  earth  being  of  unlimited  extent  and  subjected  to  no  external 


36 


NON-COHESIVE   EARTH.      GRARHICAL  METHODS 


force  but  its  own  weight.  Then  by  Art.  n,  the  earth  thrust 
on  AO  acts  parallel  to  the  free  surface  or  to  the  line  of  greatest 
declivity  5  06  5. 

The  construction  for  the  earth  thrust  caused  by  the  wedge 
of  maximum  thrust  to  the  right  of  AO  is  similar  to  that  of 
Fig.  9  and  will  be  only  briefly  indicated.  The  surface  O65  is 


H' 


divided  into  equal  parts;  then  the  weights  of  the  prisms  AObi, 
Abib2,  etc.,  will  each  equal  ^  w.p.  Obi,  where  p  =  length  of 
perpendicular  from  A  upon  5  065.  By  addition,  compute  the 
weights  of  the  trial  wedges  of  rupture  AObi,  AOb2,  etc.,  and  lay 
off  on  a  vertical  Ag  to  scale,  the  lengths  ggi,  gg2,  etc.,  to  represent 
these  successive  weights.  Then  with  A  and  g  55  Centers  and  the 
same  radius  =  AH  =  gg±,  describe  the  arcs  Hdai  and  g4Si  and 
lay  off  HAd  =  <p,  H'AH  being  horizontal.  Then,  as  before, 
lay  off  g±Si  =  dai,  g4s2  =  da2,  etc. ;  whence  gsi,  gs2,  . . . ,  make  the 
angles  <p  with  the  normals  to  planes  A bi,  Ab2,  . . . ,  and  thus 
give  the  directions  of  their  reactions.  On  drawing  through 
gi,  g2,  .  . .  ,  lines  parallel  to  5  065  to  intersections  Ci,  c2,  . . . ,  with 
gsi,  gs2,  .  .  .  ,  the  longest  of  the  lines  gid,  g2c2,  .  .  .  ,  to  the  scale  of 
force,  gives  the  thrust  on  AO.  It  is  thus  represented  by  g*Ci 


26J  THRUST   ON  A  VERTICAL  PLANE  37 

and  it  of  course  acts  parallel  to  5  Obb.  The  plane  of  rupture 
on  the  right  of  AO  corresponding,  is  Ab4. 

There  is  likewise  a  plane  of  rupture  on  the  left  of  AO,  cor- 
responding to  an  earth  thrust,  which  for  equilibrium  must  be 
equal  and  directly  opposed  to  the  thrust  on  AO  just  found. 
The  construction,  indicated  to  the  left  of  Ag,  is  similar  to  the 
preceding.  The  equal  parts  Oi,  12,  23,  . .  .  ,  were  each  made 
only  half  the  length  of  Obi,  so  that  the  successive  trial  wedges 
AOi,  A02,  .  .  .  ,  are  only  half  the  weight  of  wedges  AObi,  A0b2, 
.  .  .  ;  but  by  using  double  the  previous  force  scale,  we  can  regard 
ggi>  ££2,  •  •  •  ,  as  representing  their  weights. 

On  constructing  H'Ad'  =  <p  and  extending  Ai,  A2,  .  .  .  , 
to  meet  arc  Hdi  d'H'  at  HI,  n%,  .  .  . ,  we  next  lay  off,  #4n  =  d'ni, 
gfo  =  d'th,  g^3  =  d'ns,  .  .  .  ;  then,  gri  gr2,  gr3}  .  .  .  ,  make  the 
angles  <p  with  the  normals  to  the  planes  A  i,  ^2,  A$,  .  .  .  ,  and  are 
thus  parallel  to  their  reactions,  Art.  19. 

We  ncxw  draw  through  gi,  g2,  .  .  .  ,  lines  parallel  to  the  free 
surface  to  intersections  ei}  e2,  ...  ,  with  gr\,  gr2,  .  .  .  ;  whence  by 
the  reasoning  of  Art.  21,  gses,  the  maximum  intercept  of  the 
type  gfrj  to  the  scale  of  loads,  represents  the  active  earth  thrust 
on  AO.  Since  the  scale  of  loads  for  the  construction  to  the  left 
of  AO  was  taken  as  double  that  for  the  construction  to  the 
right  of  AO,  the  length  gse3  should  be  double  the  length  g4c4j 
which  we  find  to-  be  the  case. 

The  earth  thrust  on  AO,  due  to  the  wedge  of  rupture  AOb^A 
on  the  right,  acting  to  the  left,  parallel  to  g4c4,  is  thus  exactly 
equal  and  opposed  to  the  thrust  on  AO,  due  to  the  wedge  of 
rupture  .403^4  on  the  left,  acting  parallel  to  e3g3  and  to  the 
right,  otherwise  equilibrium  could  not  be  maintained.  Also 
by  Art.  24,  the  center  of  pressure  of  either  thrust  is  on  AO  at 
a  distance  (1/3)  AO  above  A. 

To  find  the  actual  thrusts  on  the  planes  Ai,  A2,  A^,  A$, 
draw  a  vertical  through  e3  and  extend  the  lines  g&\,  gtfz,  g4e4, 
gbeb  to  meet  it  at  h,  k,  t4,  /6;  then  the  lines  gh,  gk,  gt±,  gt$  (not 
drawn)  give  the  thrusts  on  the  planes  Ai,  A2,  A 4,  ^5.  The 
proof  is  easy.  Thus  the  weight  of  wedge  AOiA  is  ggi  to  scale; 
the  thrust  on  AO  is  given  by  gJi  =  g3es.  The  resultant  gti  of 


38  NON-COHESIVE   EARTH.      GRAPHICAL   METHODS 

these  two  forces  is  the  resultant  thrust  on  plane  Ai.     Similarly 
for  the  other  wedges. 

The  lines  gti,  gk,  gt±,  gt*>,  all  lie  above  the  corresponding 
lines  gei,  ge2,  ge±,  ge*>\  hence  the  actual  resultants  on  the  planes 
A i,  A2,  A 4,  ^5,  all  make  less  angles  than  (p  with  the  normals 
to  these  planes;  so  that  stability  is  everywhere  assured.  But 
this  would  not  be  the  case  if  we  should  arbitrarily  assume  the 
direction  of  the  thrust  on  AO  to  lie  nearer  the  vertical  than  before; 
since  then  intersections,  Ci,  C2)  .  .  .  ,  would  move  toward  AOg, 
whereas  intersections  e\,  e^  .  .  .  ,  would  move  away  from  Ag\ 
hence  the  maximum  value  of  gf{  (i  =  i,  2,  .  .  .)  would  have 
a  less  horizontal  component  than  before,  whereas  the  maximum 
value  of  giei  (i  =  i,  2,  .  .  .)  would  have  a  greater  horizontal 
component  than  was  found  in  Fig.  10.  The  active  thrust 
from  the  corresponding  wedge  of  rupture  to  the  right  of  AO 
would  consequently  be  less  than  the  thrust  pertaining  to  the 
wedge  of  rupture  on  the  left  of  AO.  Equilibrium  is  thus 
impossible. 

Similar  reasoning  would  show,  for  an  unlimited  mass  of 
earth,  that,  if  the  direction  of  the  thrust  on  AO  was  taken  nearer 
the  horizontal  than  in  Fig.  10,  equilibrium  would  be  im- 
possible. Hence  we  have  a  verification  of  Rankine's  theorem' 
of  Art.  n,  that  the  pressure  on  a  vertical  plane,  in  the 
case  of  the  unlimited  mass,  subjected  to  no  external  force 
but  its  own  weight,  acts  parallel  to  the  sloping  free 
surface. 

27.  Limiting  Plane.  Suppose  the  inner  face  of  a  retaining 
wall  to  occupy  the  position  ^5,  Fig.  10,  below  the  plane  of 
rupture  A$  on  the  left  of  AO.  On  combinir.r;  the  thrust  on 
AO,  acting  to  the  left,  parallel  to  the  sloping  surf  ate  at  C,  where 
AC  =  (1/3)  AO,  with  the  weight  of  the  wedge  OA$,  the  resultant 
on  plane  ^5  can  be  found.  Using  the  left  construction,  gg&  is 
laid  off  equal  to  the  weight  of  AO$  =  %  w.p.  0^  and  gsh  =  gse3, 
representing  the  thrust  on  AO,  is  drawn  to  the  left  parallel  to 
5065  from  g5  to  t5.  Hence  a  straight  line  gtb  (not  drawn)  to 
the  scale  of  force  gives  the  thrust  on  ,4  5,  provided  its  angle 
with  the  normal  to  A$,  marked  X  in  the  figure,  does  not  exceed 


27-29]  LIMITING  PLANE  39 

<p',  the  angle  of  friction  of  earth  on  wall.  It  acts  at  B  on  4 5, 
where  AB  =  (1/3)  ^5. 

Considering  first  the  case  where  \~<?'  and  recalling  what 
was  proved  above,  that  the  resultants  on  planes  Ai,  A2,  and 
A 4  make  less  angles  than  <p  with  the  normals  to  those  planes, 
it  is  seen  that  slipping  is  only  impending  on  the  plane  of  rupture 
A$,  if  X  <  v  or  on  both  A$  and  ^5,  if  X  =  <p'.  When  X  <  /, 
if  the  wall  moves  over  at  the  top,  the  break  will  occur  only  along 
the  plane  A3,  the  earth  between  this  plane  and  the  wall  moving 
over  with  the  wall.  If  X  =  /,  the  slipping  will  occur  along 
either  or  both  of  the  planes  A  3,  A$. 

It  is  convenient  to  give  a  name  -to  this  plane  of  rupture  on 
the  left  of  AO  (A 3  in  Fig.  10)  and  it  will  be  called  hereafter  the 
limiting  plane  for  reasons  that  will  appear  as  we  proceed. 

28.  //  the  inner  face  of  the  wall  lies  above  the  limiting  plane, 
the  construction  pertaining  to  Fig.  9  applies;  so  that  the  thrust 
on  the  wall  is  assumed  to  make  the  angle  with  its  normal,  <? 
(or  <pf  for  <p-  <  (p),  the  trial  prisms  of  rupture  are  estimated 
from  AB,  Fig.  9,  and  the  thrust  determined  as  shown  there. 

As  was  proved  in  Art.  21,  slipping  is  only  impending  along 
the  surface  of  rupture  (which  is  to  the  right  of  Ag)  and  along 
the  wall,  and  stability  is  everywhere  assured. 

When  the  inner  face  of  the  wall  is  at  the  limiting  plane  and 
V  J^.  <p,  so  that  the  thrust  on  the  wall  must  be  assumed  to  make 
the  angle  tp  with  its  normal,  the  construction  of  Fig.  9  leads  to 
the  same  result  as  the  construction  of  Fig.  10.  This  is  plain, 
since  the  direction  of  the  thrust  on  the  wall  by  either  method 
is  the  same,  and  the  amounts  must  be  the  same,  otherwise  the 
wall  reaction  by  the  one  method  could  not  balance  the  thrust 
as  given  by  the  other  method.  For  <p  <  (p,  see  Art.  29. 

29.  X  >  <p'.    In  case  the  construction  of  Fig.   10  gives  a 
thrust  on  the  wall  which  makes  a  greater  angle  with  its  normal 
than  <p' ',  then  the  direction  of  the  thrust  on  the  wall  must  be 
assumed  to  make  the  angle  /  with  its  normal  and  the  resulting 
thrust  found  by  the  construction  of  Fig.  9. 

If  for  definiteness  we  take  A  5  in  Fig.  10  as  the  inner  face  of 
the  wall  and  combine  its  reaction  (equal  and  opposed  to  the 


40  NON-COHESIVE   EARTH.      GRAPHICAL  METHODS 

earth  thrust  thus  found)  with  the  weight  of  earth  A0$,  we  find 
the  thrust,  acting  from  left  to  right,  on  the  plane  AO,  and  for 
equilibrium  this  thrust  must  be  opposed  and  equal  to  the 
earth  thrust  of  the  mass  to  the  right  of  the  plane  AO  as  found 
by  a  construction  similar  to  that  to  the  right  of  Ag  in  Fig.  10, 
only  with  the  direction  of  the  thrust  on  AO  as  just  found.  But 
this  earth  thrust  on  AO  acting  to  left  must  have  a  greater 
horizontal  component  than  the  thrust  g^c^  of  Fig.  10,  so  that 
when  combined  with  the  weight  of  A0$,  the  resultant  on  ^5 
will  make  the  angle  <?'  with  the  normal  to  ^5,  and  thus  lie 
nearer  the  horizontal  than  by  the  construction  of  Fig.  10,  since 
it  was  assumed  that  <p'  <  X. 

As  a  consequence,  the  equal  angles  ggiCi,  ggzte,  .  .  .  ,  ggsCtj 
must  increase,  for  then  the  points  c  move  away  from  the  vertical 
Ag  and  the  horizontal  component  of  the  maximum  value  of 
grcr  (r  =  i,  2,  .  .  . ,  5) — which  represents  the  thrust  on  AO  for 
the  assumed  direction — is  greater  than  the  horizontal  component 
of  g4c4.  Let  such  a  thrust  on  AO  =  T.  Then  the  resultant  on 
A$,  say,  is  found  by  combining  the  weight  of  wedge  A O$  =ggs 
(to  scale  used  for  wedges  to  left  of  AO)  with  T,  represented  by  a 
line  through  g3%  lying  above  63  and  having,  a  greater  horizontal 
component  than  g3e3.  The  resultant  on  A 3  is  thus  represented 
by  a  line  through  g,  lying  above  ges  and  making  a  less  angle  than 
<p  with  the  normal  to  ^3.  Similarly  it  can  be  shown  that  the 
thrusts  on  the  planes  Ai,  A2,  A 4,  lie  nearer  the  horizontal  than 
the  positions  g/i,  gt2,  g/4,  given  in  Fig.  10,  and  thus  make  angles 
less  than  <p  with  the  normals  to  the  planes,  so  that  the  conditions 
for  the  stability  of  the  granular  mass  are  everywhere  assured. 
The  method  of  proceeding,  for  the  case  indicated  at  the  beginning 
of  this  article,  is  thus  established.  In  this  case;' there  is  no 
plane  of  rupture  between  A O  and  ^5,  but  slipping  is  impending 
along  the  wall  face  ^5  and  along  the  plane  of  rupture  to  the 
right  of  AO,  as  found  by  the  construction  of  Fig.  9. 

Since  X  <  <p,  this  case,  X  >  <p',  can  only  occur  when  /  <  <p, 
as  when,  in  the  design  of  walls  with  a  plane  inner  face,  it  is 
arbitrarily  assumed  that  <p'  =  (1/3)  <f>,  say.  For  stepped  walls, 
where  <?'  =  </?,  the  case  cannot  occur. 


30]  SUMMARY  41 

30.  Summary.    For  all  cases  of  surface  slope,  when  the 
inner  face  of  the  wall  is  battered,  find  the  limiting  plane  by  the 
construction  of  Fig.  10  or  preferably  by  a  much  simpler  method 
to  be  given  in  Art.  65. 

(1)  When  the  inner  face  of  the  wall  makes  a  less  angle  with 
the  vertical  than  the  limiting  plane,  assume  the  direction  of  the 
thrust  on  it  as  making  the  angle  <p  (or  $   for  $   <  <p)  with  its 
normal  and  proceed  as  in  Art.  21,  referring  to  the  construction 
of  Fig.  9. 

(2)  If  the  inner  face  of  the  wall  lies  below  the  limiting  plane, 
find  by  the  method  of  Arts.  26-27  the  thrust  on  the  wall  and 
suppose  it  to  make  the  angle  X  with  the  normal  to  the  wall, 
Fig.  10.     If  X  <_  <p'}    the    thrust    on    the    wall    is    correctly 
determined. 

By  this  method,  the  thrust  on  a  vertical  plane  passing 
through  the  heel  of  the  wall  is  first  found,  the  thrust  having 
the  direction  of  the  sloping  surface.  This  thrust  is  then  com- 
bined with  the  triangular  mass  of  earth  lying  between  the 
vertical  plane  mentioned,  and  the  wall,  to  find  the  resultant 
pressure  on  it. 

(3)  When    X  >  <f>',  which   could  only  occur  when   </  <  <p, 
the  direction  of  the  thrust  on  the  wall,  will  have  t"o  be  assumed 
as  making  the  angle  <p  with  its  normal  and  the  construction  of 
Art.  21,  Fig.  9,  effected  to  find  this  thrust. 

(4)  If  the  wall  leans  toward  the  earth,  the  construction  of 
Fig.  9  directly  applies,  the  overhang  not  being  too  great  or 
when  the  inner  face  makes  an  angle  with  the  vertical  not  exceed- 
ing, say  10°.     See  Art.  62  for  the  general  case. 

31.  In  the  following  table  are  given  the  values  of  K,  0  and 
7,  corresponding  to  the  thrust,  acting  parallel  to  the  free  surface, 
on  a  vertical  plane,  for  the  commonly  used  values  of  the  angle 
of  repose,  <p  =  30°  and  <p  =  33°  41'. 

The  free  surface  is  supposed  to  rise  going  to  the  right,  as  in 
Fig.  10,  and  to  make  an  angle  i  with  the  horizontal. 

The  angle  made  by  the  plane  of  rupture  to  the  left  of  AO, 
Fig.  10,  with  the  vertical  will  be  called  0;  the  angle  made  by 
the  plane  of  rupture  on  the  right  of  AO  with  the  vertical  7;  so 


42 


NON-COHESIVE  EARTH.      GRAPHICAL  METHODS 


that  the  "  limiting  plane"  makes  the  angle  0  with  the  vertical. 
Thus  in  Fig.  10,  0  =  OA$,  7  = 


V?  =30° 

^ 

>  =  33°  41' 

* 

K 

/3 

y 

K 

0 

v 

o  

o  1667 

30°  oo' 

30°  oo' 

O.  1433 

28°  10' 

28°  10' 

5°..  , 

1686 

27°  20' 

32°  31' 

.  1447 

26°  09' 

30°  10' 

10° 

174.8 

24.°  <;o' 

ac°  jo 

IAQT. 

24°  03' 

32°  16' 

ic° 

1865 

21°  SS' 

^8°  os' 

1^77 

21°  46' 

34°  33' 

20°:  * 

2O7I 

18°  25' 

4.1°  15 

I7l8 

19°  08' 

o<t  oo 
T7  II 

25°. 

2468 

iV  38' 

46°  22' 

IQCQ 

15°  Si' 

40°  28' 

30° 

A^O 

0° 

60°  oo' 

24^1 

10°  so' 

45°  20' 

33°  4i'.' 

4l6l 

0° 

S6°  19' 

The  quantity  K  is  the  factor  in  the  formula 
E  =  K  w  h2 

and  is  the  same  whether  the  surface  rises  or  falls  in  going  to 
the  right,  the  thrust  E  always  acting  parallel  to  the  free  surface. 
It  was  computed  from  Rankine's  formula,  Art.  48,  and  checked 
by  the  neat  construction,  Art.  58,  Fig.  32. 

The  values  of  0  and  7  were  computed  very  easily  from  the 
formulas  of  Art.  65,  and  were  checked  by  use  of  Fig.  37,  Art.  65. 

From  considerations  of  symmetry,  when  the  free  surface  50^5, 
Fig.  10,  is  horizontal  or  i  =  o,  we  must  have  0  =  7;  but  it 
requires  an  analytical  demonstration  to  show,  when  i  —  <p,  that 
0  =  o,  7  =  90°  —  <p,  or  strictly  that  as  i  approaches  <p  in- 
definitely, the  limiting  plane  approaches  the  vertical,  and  the 
plane  of  rupture  to  the  right  of  AO  approaches  indefinitely 
the  natural  slope. 

It  may  be  observed  also,  what  is  obvious  from  the  figure, 
that  when  the  free  surface  falls  to  the  rigut,  0  and  7  are 
interchanged.  From  the  large  values  of  7  (or  the  small  values 
of  0)  given  in  the  table  for  values  of  i  near  <pt  it  is  seen  that  the 
construction  of  Fig.  10  is  impracticable  for  such  large  values  of 
i',  hence  the  results  of  a  subsequent  chapter  are  anticipated  to 
aid  in  immediate  practical  computations. 

Ex.  I.  Find,  by  aid  of  the  table,  the  Rankine  thrust  in  pounds,  E  = 
K  w  h?,  when  i  =  20°,  <f>  =  30°,  w  =  100  Ib.  per  cu.  ft.  and  h  =  20  feet. 


31,  32] 


SURCHARGED   WALL 


43 


Ex.  2.  Show  by  aid  of  a  construction  similar  to  that  pertaining  to  the 
wedges  to  the  right  of  AO  in  Fig.  10,  that  when  i  =  o,  <p  =  45°,  the  coefficient 
K  =  0.0858,  the  thrust  being  taken  horizontal  on  a  vertical  plane. 

32.  Surcharged  Wall.  Whenever  the  earth  behind  a  re- 
taining wall  extends  above  its  top,  the  wall  is  said  to  be  sur- 
charged, and  the  portion  of  earth  above  the  level  of  the  top  of 


FIG.  ii 

the  wall  is  called  the  surcharge.  In  Fig.  n  is  shown  a  sur- 
charged wall,  the  surcharge  sloping  up  to  61,  the  surface  bib* 
being  then  taken  horizontal. 

As  usual,  one  foot  length  of  wall  is  considered,  ACDF  being 
its  medial  section  and  B  the  point  where  AC  meets  the  sloping 
surface.  The  earth  thrust  E  will  be  assumed  to  make  the 
angle  <?'  with  the  normal  to  AC,  where  <?'  <.  <p.  For  convenience 
in  computation,  lay  off  along  the  level  part  of  the  surface  the 
equal  lengths,  bj)2,  b2bS)  .  .  .,  and  draw  the  trial  planes  of  rupture 
Abi,  Abz,  ....  On  drawing  BbQ  parallel  to  the  base  Abi,  to 
intersection  bQ  with  b$i  produced,  we  have,  area  triangle  ABbi  = 
area  triangle  Abjbi.  Also  letting  p  =  Af  =  perpendicular  from 
A  on  &o&4,  we  have,  areas  of  triangles  Abjb\t  Abib2,  Abzbz,  •  -  -  , 
equal  to  X  P-  Wi,  X  P-  bJ>*,  •  •  •  5  whence  the  weights  of  the 
successive  prisms, 

,  ABb,bz,  . 


44  NON-COHESIVE   EARTH.      GRAPHICAL  METHODS 

are  equal  to  >2  w  p  multiplied  by 


Here,  w  =  weight  of  i  cu.  ft.  of  earth  in  pounds,  p  =  Af  in 
feet,  therefore  the  weights  of  the  successive  trial  prisms  (which 
are  quickly  obtained  by  addition)  are  given  in  pounds.  We 
next  lay  off  on  a  vertical  through  A,  from  any  convenient  point 
g>  ££i>  ££25  ggsj  .  .  .  ,  to  the  force  scale,  equal  to  these  successive 
weights. 

The  construction  now  proceeds  as  in  Art.  21,  Fig.  9.  Thus 
AH  being  horizontal,  construct  HAd  —  $  and  with  A  and  g 
as  centers  and  a  common  radius  Ag,  describe  the  arcs  gdid  and 
A$I.  Lay  off  Asi  =  dai,  As2  =  Ja2,  .  .  .  ,  and  draw  gsi,  gsz,  .  .  .  , 
which  thus  make  the  angles  <p  with  the  normals  to  planes  Abi, 
Ab2,  .  .  .  ,  Art.  19,  and  give  the  directions  of  their  reactions. 
Lines  are  drawn  through  gi,  g2}  .  .  .  ,  parallel  to  the  assumed 
direction  of  E,  to  intersections  Ci,  c2,  .  .  .  ,  with  gsi,  gs2,  .  .  .  ; 
whence  the  longest  of  the  intercepts  of  the  type  g^,  which  is 
gzcz,  to  the  force  scale,  gives  the  thrust  E  in  pounds.  Of  course 
other  trial  planes  of  rupture  must  be  drawn  near  Ab2  and  the 
force  triangles  corresponding,  drawn  to  obtain  the  thrust  more 
accurately  and  it  is  further  advisable  .to  use  as  large  a  scale, 
both  for  distance  and  force,  as  is  convenient. 

The  limiting  plane  for  this  figure  cannot  be  accurately  found 
since  the  direction  of  the  thrust  on  the  vertical  plane  through 
A  for  the  unlimited  mass  is  unknown.  This  direction  may  be 
assumed  roughly  as  parallel  to  Bb2  (when  Ab2  is  the  plane  of 
rupture),  and  the  limiting  plane  found  as  indicated  in  Art.  31. 
As  an  illustration,  suppose  in  Fig.  n  that  Bb2  rriakes  an  angle 
of  20°  with  the  horizontal  and  that  <p  =  30°;  tnen  the  table 
of  Art.  31  can  be  utilized  by  entering  it  with  i  =  20°,  <p  =  30°, 
whence  /3  =  18°  25'.  The  directions  of  Art.  30  can  then  be 
followed.  Thus,  if  AC  lies  above  the  limiting  plane,  the  thrust 
is  given  correctly  by  the  construction  of  Fig.  n;  if  below  it, 
then  assuming  the  direction  of  the  thrust  on  the  vertical  plane 
through  A  as  having  the  direction  Bb2,  find  its  amount  by  a 
construction  analogous  to  that  of  Fig.  n,  only  the  point  B  is 


32,  33]  SURCHARGED   WALL  45 

now  to  be  taken  where  the  vertical  through  A  intersects  the 
free  surface.  The  thrust  thus  found  is  to  be  combined  with 
the  weight  of  earth  lying  between  the  vertical  plane  through 
A  and  the  wall  to  get  the  thrust  on  the  wall,  provided  its  obliquity 
X.^L</.  When  X  ><?',  the  construction  of  Fig.  n  directly 
applies.* 

Remark.  When  the  earth  contour  is  irregular  or  curved, 
lay  off  ggi,  gg2,  .  .  .  ,  equal  to  the  weights  of  the  successive  trial 
wedges  of  rupture  and  proceed  as  before. 

Center  of  Pressure.  A  simpler  construction  for  the  thrust 
on  a  surcharged  wall  will  be  given  in  Art.  39,  and  a  resulting 
computation  in  Art.  42,  to  find  the  center  of  pressure  for  various 
heights  of  surcharge.  See  the  table  giving  the  results. 

33.  The  disturbing  influence  of  the  wall  friction  in  changing 
the  usual  distribution  of  the  stresses  in  an  unlimited  mass  of 
earth  can  be  illustrated  by  reference  to  Fig.  10.  Thus  if  ^5 
is  a  plane  in  an  unlimited  mass  of  'earth  with  a  plane  surface, 
the  thrust  on  the  vertical  plane  AO  acts  parallel  to  the  surface 
at  Cj  where  AC  =  (1/3)  AO  and  meets  ^.5  at  B,  where  AB  = 
(1/3)  ^5,  through  which  point  the  resultant  weight  of  A0$  also 
acts.  The  resultant  pressure  on  the  plane  A$  thus  acts  at  the 
third  point  on  ^.5  and  therefore  corresponds  to  a  pressure  uni- 
formly increasing  from  5  to  A,  Art.  15. 

A  similar  conclusion  holds  for  any  plane,  Ab2  or  Ab$  to  the 
right  of  AO.  Now  suppose  a  wall  introduced  and  that  the 
wall  friction  causes  the  thrust  on  AO  to  lie  above  or  below 
the  direction  first  assumed,  then  the  resultant  pressure  on  the 
plane  of  rupture,  say  Abt,  will  not  pass  through  its  third  point, 


*In  this  last  case  a  simpler  test  is  available.  Thus  having  made  the  con- 
struction, Fig.  u,  lay  off  from  g  downward  to  scale,  gh,  representing  the 
weight  of  the  earth  between  the  vertical  plane  through  A  and  the  wall,  and 
draw  from  h  to  the  right  a  (dotted)  line  to  t,  equal  and  parallel  to  g2c2,  the 
reaction  E  of  the  wall.  Then  tg  is  the  thrust  on  the  vertical  plane  through  A. 
If  the  angle  it  makes  to  the  right,  with  the  upward  direction  of  Ag,  is  greater 
than  the  corresponding  angle  made  by  Bb2  with  Ag,  then,  by  the  reasoning  of 
Art.  29,  the  construction  of  Fig.  n  is  the  true  one  and  slipping  will  only  be 
impending  along  Ab2  and  A  B.  This  test  avoids  the  construction  of  the 
limiting  plane. 


46  NON-COHESIVE   EARTH.      GRAPHICAL   METHODS 

so  that  the  pressure  on  the  plane,  which  must  be  zero  at  the 
surface,  is  no  longer  uniformly  increasing.  But  at  a  little 
distance  from  the  wall,  the  distance  probably  increasing  with 
the  depth,  the  usual  direction  of  the  thrust  in  an  unlimited 
mass  is  doubtless  attained  and  the  pressures  along  any  plane 
are  uniformly  increasing  from  the  surface  down.  The  incon- 
sistency is  due,  in  part,  to  having  assumed  a  plane  surface  of 
rupture,  for  the  case  in  question,  whereas  experiment  shows  it 
to  be  a  curved  surface.  However,  for  clean  sand  having  little 
cohesion,  the  curvature  is  only  slight  and  the  hypothesis  of  a 
plane  surface  should  lead  to  results  that  are  approximately  true, 
which  experiments  on  retaining  walls  verify. 

The  sHding-wedge  theory  is  due  to  Coulomb  and  it  has  been 
developed  by  a  number  of  authors  whose  results  vary  only  as 
to  the  direction  assumed  for  the  thrust  against  the  wall.  Many 
European  authors  assume  the  thrust  as  making  the  angle  <p 
with  the  normal  to  the  wall,  thus  including  the  wall  friction, 
whereas  other  writers  follow  Rankine  in  taking  the  thrust  on  a 
vertical  plane  as  parallel  to  the  plane  surface  assumed.  Ran- 
kine developed  his  theory  from  certain  general  considerations 
affecting  the  relations  of  stresses  in  an  unlimited  mass  of  earth. 
In  reality  he  uses  an  infinitesimal  wedge  of  rupture,  as  we  shall 
see  in  Chapter  III,  and  by  integration  finds  the  thrust  on  a  vertical 
plane.  For  the  kind  of  mass  supposed,  not  subjected  to  any 
external  force  (like  wall  friction)  but  its  own  weight,  his  results 
are  exact,  on  the  hypothesis  of  earth  endowed  with  friction  but 
without  cohesion.  It  is  only  when  the  application  of  the  Rankine 
theory  to  retaining  walls  is  made,  ignoring  the  wall  friction,  that 
objection  can  be  made.  Two  recent  author^  Boussinesq  and 
Resal,  have  endeavored  to  complete  the  Rankine  theory  by 
considering  the  influence  of  the  wall  friction  in  modifying  it  in 
the  vicinity  of  the  wall.  In  Resal's  "Poussee  Des  Terres,"  I, 
the  assumption  is  made  that  the  earth  pressures  on  any  plane 
in  a  mass  of  earth  increase  uniformly  in  going  from  the  surface 
downward  and  that  the  thrust  on  a  wall  has  an  obliquity  <p. 
After  an  intricate  analysis  leading  to  complicated  equations,  a 
method  of  finite  differences  has  to  be  eventually  invoked  to 


33,  34]  CENTERS    OF   GRAVITY  47 

reach  numerical  results,  which  are  tabulated  in  excellent  shape 
for  use  in  practice,  particularly  in  the  second  volume.  As  the 
first  part  of  his  assumption  as  to  uniformly  increasing  pressures, 
although  plausible,  is  open  to  question,  his  theory  can  scarcely 
claim  to  be  exact.  It  is,  however,  the  most  consistent  theory  of 
all,  and  it  is  a  regret  that  it  is  too  complicated  to  present  even 
in  outline  in  an  elementary  treatise.* 

It  is  evident  that  in  an  unlimited  mass  of  earth,  the  pressures 
along  any  inclined  plane  must  increase  uniformly  from  the 
surface  down,  as  for  the  case  of  the  vertical  plane,  Art.  n.  The 
plane  surface  of  rupture  leads  to  such  a  state  of  stress,  as  seen  in 
Fig.  10 ;  hence,  for  the  unlimited  mass,  the  surface  of  rupture  is 
plane  and  the  assumption  leads  to  exact  results.  In  fact  it 
leads  to  the  Rankine  formula,  as  will  be  shown  in  Art.  48. 

One  other  remark  may  be  pertinent.  If  we  suppose  a  perfectly 
rigid  vertical  wall,  lying  above  the  limiting  plane,  on  a  rigid 
base,  backed  by  incompressible  earth  level  with  its  top,  then 
(for  these  impossible  conditions)  there  will  be  no  wall  friction 
brought  into  play  and  the  Rankine  thrust  against  the  wall  will 
be  exerted.  If,  however,  the  wall  and  foundation  are  elastic, 
the  top  will  move  over,  wall  friction  is  experienced  and  the 
horizontal  component  of  the  thrust  becomes  less  than  before, 
Art.  22.  Now  a  little  ways  in  the  interior,  the  Rankine  thrust 
is  still  exerted.  What  becomes  of  this  excess  in  horizontal 
thrust?  It  must  necessarily  be  transmitted  to  the  ground  and 
resisted  by  friction  along  the  ground  surface;  which  it  is  generally 
perfectly  capable  of  doing.  The  state  of  stress  is  complicated 
and  its  solution  is  attempted  in  Resal's  treatise  quoted  above. 

34.  Centers  of  Gravity.  In  Fig.  12,  let  the  trapezoid  A  BCD 
represent  the  medial  section  of  a  wall.  Its  center  of  gravity 
will  coincide  with  the  (so  called)  center  of  gravity  of  the  trape- 
zoid. Let  DC  =  a  and  AB  =  b,  be  the  lengths  of  the  parallel 


*  One  of  Resal's  conclusions,  however,  may  be  stated  since  it  agrees  with 
a  theorem  otherwise  demonstrated  above :  that  when  the  inner  face  of  the  wall 
lies  at  or  below  the  "limiting  plane,"  the  thrust  on  it  is  to  be  found  as  explained 
in  Art.  27,  taking  <p'  =  <f>.  Thus  Fig.  10,  if  the  wall  A  5  lies  at  or  below  the 
plane  of  rupture  A 3,  the  thrust  on  AO  acting  parallel  to  the  surface  is  found 
and  then  combined  with  the  weight  of  earth  AO$  to  obtain  the  thrust  on  ^5. 


48  NON-COHESIVE   EARTH.      GRAPHICAL  METHODS 

sides,  whose  mid-points  are  /  and  N.  Then  the  center  of 
gravity  lies  on  the  medial  IN.  It  will  now  be  proved  that  if 
DC  be  produced  to  F  and  BA  to  E,  making  CF  =  b  =  AB  and 
EA  =  a  =  DC,  then  the  straight  line  EF  will  cut  NI  at  G, 
the  center  of  gravity  of  the  trapezoid.  Let  y  —  perpendicular 


FIG.  12 


distance  from  G  to  AB,  and  h  =  perpendicular  distance  from 
C  to  AB.  Divide  the  figure  into  two  triangles  by  a  line  from 
A  to  C  and  take  moments  about  A, 

a  +  b  .          bh  h   .   ah  2  , 
.'.-      -h     =  —--  ---  h 


2323 

(b  +  20)  h 


Now  if  G  is  the  center  of  gravity,  lying  somewhere  on  NI,  we 
have, 

y  _  NG  _    b+2a 

It  ~  ~NI  ~  30  +  3&  ' 
By  geometric  division, 

NG'       _  NG  _  b  +  2a  _  b/2  +  a  _    EN 
NI  -  NG  ~"Gl"  2b  +  a  ~  b  +  'ajV  JF  ' 

since  we  laid  off,  EA  =  a  and  CF  =  b. 

Hence  from  the  equality  of  the  second  and  fifth  ratios,  it 
is  seen  that  EF  cuts  NI  at  G,  the  center  of  gravity  of  the  trape- 
zoid, the  two  triangles  ENG  and  GIF  being  similar  and  giving 
the  above  proportion.  The  construction  indicated  above  is 
thus  justified. 


34,35] 


CENTERS   OF   GRAVITY 


49 


Regarding  AB  as  horizontal,  let  x  represent  the  horizontal 
distance  from  A,  Fig.  12,  to  the  vertical  through  the  center  of 
gravity  G  of  the  trapezoid,  when  CB  is  inclined  at  an  angle  a 
with  the  vertical.  Dividing  A  BCD  into  the  two  triangles 
ADC  and  ACB  and  taking  moments  about  A,  we  derive  after 
reduction, 

_  _  2b2  +  2ab  -  a2  -  (20,  +  ft)  h  tan  a 

3  (a  +  ft) 

When  the  projection  of  C  on  AB  falls  to  the  right  of  B,  change 
the  sign  of  tan  a. 

The  center  of  gravity  of  any  quadrilateral  MNOP,  Fig.  13, 
can  be  found  as  follows  :  Let  E  be  the  mid-point  of  MO,  EA  = 
1/3  EP,  EB  =  1/3  EN;  join  B  and  A  with  a  straight  line  and 
let  it  intersect  MO  at  I;  if  now  on  BA,  BG  is  laid  off  equal  to  I  A, 
tlien  G  is  the  center  of  gravity  of  the  quadrilateral  MNOP.  Noting 
that  A  and  B  are  the  centers  of  gravity  of  the  triangles  MOP 
:and  MNO  and  that  PN\\AB, 


area  MNO      GA 


I  A       HP      area  MOP      GB 

being  regarded  as  the  position  of  the  resultant  of  two  forces 
proportional  to  areas  MNO,  MOP,  and  applied  at  B  and  A 
respectively. 

By  composition, 


BI  +  IA       BG+  GA 


IA 


BG 


BG  =  IA. 


shows   that   the   construction  indicated   is   valid.     It  is 
)ften  the  best  construction  for  the  trapezoid,  the  lines  to  be 
twn  all  lying  within  the  figure. 
35.  Test  of  the  Stability  of  Retaining  Walls. 

Ex.  i.     Consider  a  wall  of  the  trapezoidal  type,  Fig.  12,  backed  by  earth 
sloping  upward  from  C  at  an  angle  of  6°  with  the  horizontal,  the  earth  weigh- 
ig  loo  and  the  wall  140  Ib.  per  cu.  ft.     Let  a  =  2',  b  =  10',  h  =  20'  and 
=  17°  2i',_whence  horizontal  projection  of  CB  =  h_tan  a  =  6.25'.     By  the 
formula  for  x  or  the  construction  of  Art.  34,  we  find  x  =  4.12'. 


50 


NON-COHESIVE  EARTH.      GRAPHICAL  METHODS 


The  earth  thrust  against  CB  for  <p  =  33°  41',  has  been  found,  Ex.  I,  Art. 
23,  to  be  E  =  9550  Ibs.  It  acts  at  (1/3)  BC  above  B  on  BC  and  makes  an 
angle  <p  =  33°  41'  with  the  normal  to  BC.  Hence,  as  in  Fig.  n,  lay  off  this 
direction  at  the  point  of  action  on  the  wall  and  extend  it  to  meet  the  vertical 
through  the  center  of  gravity  of  the  wall,  which  passes  x  =  4.12'  to  the  right 
of  the  outer  toe.  At  the  point  of  intersection,  combine  the  weight  of  the 
wall,  W  =  16,800  Ibs.  acting  vertically  downward,  with  the  earth  thrust 
E  =  9550  Ibs.,  and  find  that  the  resultant  cuts  the  base  3.6'  from  the  outer 
toe  or  1.4'  to  the  left  of  its  center.  It  is  often  best  to  combine  the  forces  as 
illustrated  in  Fig.  6,  if  thereby  a  larger  scale  can  be  used. 

The  resultant  on  the  base  is  found  to  make  an  angle  of  14°  09'  with  the 
vertical;  hence,  by  Art.  16,  the  factor  of  safety  against  sliding  is, 


tan  0 


tan  14°  09' 


tan  6' 
•25 


where  6"  is  the  angle  of  friction  of  masonry  on  earth.  If  tan  B"  is  taken  as 
0.33  for  moist  or  wet  clay  and  0.50  for  dry  clay,  the  factors  of  safety  are, 
1.3  and  2  respectively. 

The  vertical  component  of  the  resultant  on  the  base  is  found  to  be  F  — 
24,200  Ibs.;  hence  the  soil  reactions  at  the  toes  are,  by  formula  (3)  of  Art.  15, 


10 


10 


or  4450  Ib.  per  sq.  ft.  at  the  outer  toe  and  390  Ib.  per  sq.  ft.  at  the  inner  toe, 
which  are  safe  values  for  dry  sand  or  dry  clay.  The  wall  is  thus  stable  for 
a  fairly  good  earth  foundation  and  the  entire  base  is  under  compression. 
Since,  in  this  example,  a.  =  17°  21'  and,  from  the  table  in  Art.  31,  /3  >  26°, 
BC  in  Fig.  12  lies  above  the  "limiting  plane  "*and  the  earth  thrust  is  correctly 
estimated  by  the  construction  of  Fig.  9,  where  the  full  friction  between  the 
earth  and  wall  is  allowed. 

Ex.  2.  Investigation  of  the  stability  of  the  wall  of  Ex.  i  by  the  Rankine 
method.  The  wall  A  BCD  is  drawn  to  scale  in  Fig.  14  (the  original  scale  was 
2  ft.  to  i  in.).  AB  =  10',  CD  =  2',  height  of  wall  BL  =  20',  CL  =  6.25', 
vertical  BH  =  20.65  ft. 

The  earth  thrust  on  the  vertical  plane  BH,  from  the  inner  toe  to  inter- 
section with  the  free  surface  CH,  can  be  found  by  a  construction  similar  to 
that  of  Fig.  10  (using  either  the  construction  to  the  right  of  A  O  or  the  one 
to  its  left)  and  is  found  to  be  E  =  6200  Ibs.,  this  thrust  ac4J^  at  1/3  BH  above 
B,  parallel  to  HC.  Its  value  may  be  checked  by  aid  of  the  tabular  results 
in  Art.  31,  where,  for  i  =  6°,  <f>  =  33°  41',  we  find  by  interpolation,  K  = 
0.145;  whence, 

E  =  Kw.  BH2  =  0.145  X  100  X  (20.65)2  =  6180  Ibs. 

This  thrust  will  now  be  combined  with  the  combined  weight  of  the  wall 
and  of  the  earth  BCH.  To  find  x,  the  horizontal  distance  from  A  to  the 
vertical  through  the  center  of  gravity  of  these  weights,  take  moments  about  A  . 

The  weight  of  earth  BCH  =  tf  Xioo  X  20.6  X  6.25  =  6436  Ibs.  and 
it  acts  1/3(6.25)  =  2.08'  from  BH  or  7.92'  from  A\  hence  its  moment  about 


35] 


STABILITY   OF   RETAINING   WALLS 


51 


A  is  6436  X  7-92  =  50.973  ft.  Ibs.  From  Ex.  i,  the  moment  of  the  weight 
of  wall,  16,800  Ibs.,  about  A  is,  16,800  X  4.12  =  69,216  ft.  Ibs.  The  sum 
of  these  moments,  120,189,  divided  by  th.e  weight  of  wall  and  earth,  23,236, 
gives  x  =  5.18  ft.  Hence  draw  a  vertical  5.18'  to  right  of  A  to  meet  E,  the 
earth  thrust  on  BH,  at  G.  Lay  off,  to  scale,  E  =  GM  =  6200  Ibs.,  GM  being 
parallel  to  the  surface  and  passing  through  the  point  on  BH,  (1/3)  BH  above 
B,  At  M,  lay  off  MN  vertically  equaj  to  23,236  Ibs.;  extend  the  resultant 
NG  to  the  base  at  /.  A I  is  found  to  be  3.5  ft.,  or  nearly  the  same  as  before. 
The  point  I  is  still  within  the  middle  third  of  A  B,  so  that  the  whole  of  the  base 
is  under  compression  and  the  conclusions  pertaining  to  Ex.  i  obtain  here. 

The  reason  the  two  solutions,  one  by  the  wall  friction  method,  the  other 
after  Rankine,  so  nearly  agree  for  the  wall  examined,  is  because  the  thrust 


A     l 


FIG.  14 


E  on  BH,  by  the  Rankine  method,  is  combined  with  such  a  large  weight  of 
earth  BCH,  to  get  the  thrust  on  BC.  The  resultant  on  BC  thus  makes  nearly 
the  same  angle  with  the  normal  to  BC  as  in  the  previous  solution.  But  when 
BC  is  vertical,  the  two  theories  differ  widely,  the  first  assuming  the  thrust  on 
the  wall  to  make  an  angle  <p  (when  <p'  =  <p)  with  the  horizontal,  whereas 
Rankine  assumes  the  thrust  on  it,  to  act  parallel  to  CH. 

Ex.  3.  Let  Fig.  15  represent  a  portion  of  a  vertical  brick  wall  weighing 
125  Ib.  per  cu.  ft.,  the  wall  being  10'  high  and  10/3  =  3.33'  thick  and  backed 
by  earth  level  with  its  top,  the  earth  weighing  100  Ib.  per  cu.  ft.  Assume 
<p  =  tp'  =  33°4J/;  then  by  the  method  of  Art.  21,  including  all  the  wall 
friction,  we  find  E  =  K  w  h2  =  0.1304  X  100  X  io2  =  1300  Ibs. 

This  thrust  acts  at  M,  Fig.  15,  3.33'  above  B  and  is  inclined  to  the  normal 
to  BM  at  the  angle  <p.  Extend  it  to  meet  the  vertical  through  the  center  of 
the  base  C  at  G  and  combine  it  at  G  with  the  weight  of  the  wall,  4167  Ibs., 
which  acts  along  GC.  The  resultant  GI  cuts  the  base  at  7,  0.5  ft.  to  the  left 
of  its  center  and  0.06  ft.  within  the  middle  third  limit. 

Ex.  4.  Solve  the  same  example  by  the  Rankine  method.  The  thrust 
E  at  M  is  now  supposed  to  act  horizontally.  Its  amount,  using  K  =  0.1433 
from  Art.  31,  is  0.1433  X  100  X  io2  =  1433  Ibs.  At  H,  where  it  meets  the 
vertical  CH,  combine  with  the  weight  of  the  wall.  The  resultant  (dotted 
line)  is  found  to  cut  the  base  at  R,  0.52  ft.  from  A,  outside  of  the  middle  third, 
so  that  only  3  X  0.52  =  1.56  ft. — less  than  one-half  the  base — is  under  stress. 


52  NON-COHESIVE  EARTH.      GRAPHICAL  METHODS 

If  the  Rankine  theory  was  applicable,  the  wall  should  at  once  be  declared 
unsafe.  As  a  matter  of  fact  it  is  entirely  safe,  for  Sir  Benjamin  Baker 
tells  us*  that  hundreds  of  brick  revetments  have  been  built  by  the  Royal 
Engineer  officers  with  a  thickness  of  only  0.32^  for  a  vertical  wall,  which  are 
stable. 

Rankine  states,  f  in  effect,  that  (using  his  theory)  in  practice,  the  resultant 
on  the  base  is  allowed  to  cut  it  1/8  the  thickness  from  its  outer  toe  by  British 
engineers  and  1/5  the  thickness  by  French  engineers.  It  would  follow  that 
only  3/8  of  the  base  should  be  under  compression  in  the  first  instance  and 
only  3/5  in  the  second,  Art.  15. 

No  stronger  indictment  could  be  brought  against  the  application  of  his 
theory  to  vertical  rectangular  retaining  walls  than  Rankine  himself  furnishes 
us,  for  it  is  unreasonable  to  suppose  that  walls  that  have  been  stable  for 
years  should  bear  only  upon  about  half  their  base. 

The  defect  in  the  Rankine  theory  when  applied  to  rectangular  walls 
backed  with  level-topped  earth  is  that  all  the  wall  friction  is  ignored.  It  leads 
to  exaggerated  dimensions  when  the  resultant  on  the  base  is  required  to  pass 
within  its  middle  third.  A  certain  amount  of  friction  between  the  earth  and 
wall  is  always  exerted.  It  is  only  a  question  of  how  much  should  be  allowed 
in  designing  a  wall.  In  Art.  14,  suggestions  are  made  along  this  line,  which 
are  carried  out  in  the  design  of  walls,  Chapter  IV. 

These  suggestions  refer  particularly  to  walls  with  plane  inner  faces,  where 
it  was  feared  that  the  lubrication  of  the  wall  by  water  might  diminish  ma- 
terially, at  the  time  of  heavy  rains,  the  value  of  <pr.  But  when  the  rear  face 
is  stepped,  the  surface  of  impending  slipping  will  be  a  plane  in  the  earth,  touch- 
ing the  edges  of  the  steps,  and  there  seems  no  more  reason  why  the  coefficient 
of  friction  (tan  <p)  should  diminish  along  this  plane  than  along  the  surface  of 
rupture.  For  walls  exposed  to  no  vibration,  the  full  friction  of  earth  on 
earth  can  be  allowed  for  the  stepped  walls,  but  where  vibration  is  excessive, 
the  suggestion  of  Art.  14  may  be  carried  out  or  the  judgment  of  the  engineer 
invoked  to  arbitrarily  increase  the  thickness  of  the  wall  over  computed  values. 

In  this  article,  the  thrusts  are  estimated  for  usual  conditions  (no  factor 
of  safety  having  been  introduced)  which  doubtless  obtain  nearly  all  the  time. 
The  complete  design  requires  consideration,  too,  of  the  exceptional  conditions, 
as  to  which  see  Chapter  IV. 

Ex.  5.  Let  us  assume  again  the  data  of  Ex.  i,  only  with  the  surface 
CH  sloping  at  the  angle  of  repose  and  with  the  position  of  the  point  A  to  be 
determined,  so  that  the  resultant  on  the  base  AB  shall  ^at  it  (1/3)  AB  from 
A,  Fig.  16,  p.  54.  By  Art.  31,  the  "limiting  plane"  coincide^  with  BH,  so  that 
the  Rankine  method  outlined  in  Ex.  2  is  applicable  and  is  the  only  correct 
one  to  follow. 

Find  by  measurement  on  a  large  scale  drawing  or  by  computation,  BH  = 
24.17.  From  the  table,  Art.  31,  K  =  0.416  .'.  the  thrust  on  BH,  acting 
parallel  to  CH,  is  E  =  K  w  h2  =  0.416  X  100  X  (24.17)2  =  24,300  Ibs. 
Draw  a  vertical  through  D  meeting  AB  at  V,  whence  VB  =  8.25  ft.;  also 


*  Minutes  of  Proceedings,  Inst.  C.  E.,  Vol.  LXV,  p.  140;  reprinted  in  Van 
Nostrand's  Science  Series,  No.  56. 
t  Applied  Mechanics,  p.  227. 


35,  36]  PASSIVE   THRUST  53 

show,  as  in  Ex.  2,  that  the  combined  weight  of  the  portion  of  the  wall  VBCD 
and  the  earth  BCH  is  21,900  Ibs.  and  that  its  resultant  acts  through  i,  4.02' 
to  right  of  V. 

The  thrust  E  acts  on  BH,  (1/3)  BH  above  B.  Lay  off  OP  parallel  to  CH 
to  represent  this  thrust  and  draw  PQ  vertically  equal  to  the  combined  weight, 
21,900  Ibs.,  OQ  being  their  resultant.  Produce  E  to  meet  the  vertical  through 
the  center  of  gravity  of  VBCD  and  CBH  at  i  and  draw  12  parallel  to  OQ  to 
give  the  position  of  this  resultant,  OQ.  We  now  resort  to  the  method  of 
trial  and  error  to  fix  the  position  of  A.  A  V  was  first  assumed  4  ft.  and,  by  a 
construction  similar  to  the  one  that  follows,  was  found  to  be  too  large.  Tak- 
ing A  V  =  3  ft.  we  find  weight  of  ADV  =  X  X  20  X  3  X  140  =  4200  Ibs. 
and  it  acts  along  the  vertical,  (>^)  A  V  —  i  ft.  to  the  left  of  V.  This  vertical 
meets  the  line  of  action  12  of  the  resultant  whose  magnitude  is  OQ  at  2.  At 
this  point  combine  the  two  forces.  On  laying  off  QR  =  weight  of  ADV, 
OR  represents  the  magnitude  and  direction  of  the  final  resultant  on  the  base. 
Its  position  is  found  by  drawing  a  line  through  2  parallel  to  OR.  It  meets 
AB,  3.83  ft.  to  the  right  of  A.  Since  ^4.8/3  =  11.25/3  =  3.75  ft.,  it  is  seen 
that  the  resultant  passes  0.08  ft.  to  the  right  of  the  outer  third  point  or  slightly 
inside  the  middle  third  of  the  base.  The  requirement  originally  stated  is 
practically  fulfilled,  though  another  trial  might  prove  instructive  to  the 
student  as  showing  the  ease  of  working  by  this  tentative  method. 

Note.  In  practice,  walls  are  usually  designed  either  for  a  horizontal  free 
surface  or  for  one  sloping  at  the  angle  of  repose.  Supposing  <p' =  <p  (always 
true  for  a  stepped  inner  face),  then  as  shown  in  Ex.  5,  for  the  latter  case, 
the  Rankine  method  gives  the  true  solution;  but  for  vertical  walls  with 
the  earth  surface  horizontal,  a  comparison  of  the  results  of  Exs.  3  and  4  shows 
that  the  Rankine  method  is  inapplicable  and  leads  to  exaggerated  dimen- 
sions. However,  when  the  inner  face  of  the  wall  has  a  considerable  batter, 
as  usual,  it  is  seen  from  the  results  of  Exs.  i  and  2,  for  earth  surface  level  or 
nearly  so,  that  the  Rankine  method  nearly  agrees  with  the  wall-friction 
method  and  leads  to  a  slightly  greater  thickness  at  the  base,  which  is  on  the 
side  of  safety.  For  reinforced  T  walls  or  those  with  counterforts,  where  the 
earth  thrust  has  to  be  estimated  on  a  vertical  plane  a  little  distance  from 
the  vertical  slab,  the  Rankine  method  is  again  applicable;  so  that  in  practical 
design,  the  Rankine  thrust  is  called  for  in  the  majority  of  cases.  On  account 
of  its  theoretical  and  practical  importance,  it  will  be  considered  in  detail  in 
a  subsequent  chapter. 

It  is  well  at  this  stage  for  the  student  to  refer  to  the  results  of  experiments 
on  model  retaining  walls,  given  in  Appendix  II,  as  they  will  tend  to  give  him 
some  confidence  in  the  theories  advanced  above. 


36.  Passive  Thrust  against  a  Wall.  See  Art.  20.  In  Fig. 
17,  let  Ag  represent  the  vertical  inner  face  of  a  wall,  the  earth 
to  the  right  having  a  plane  surface  £123...,  and  suppose  the 
wall  subjected  to  a  force  acting  to  the  right  so  great  that  mo- 
tion is  impending  of  a  wedge  of  earth  up  some  plane  as  -42,  ^3, 


54 


NON-COHESIVE   EARTH.      GRAPHICAL   METHODS 


etc.,  and  up  the  wall;  then  the  frictional  components  of  the 
reactions  of  both  plane  and  wall  will  act  down  or  opposed  to 
the  impending  motion.  The  direction  of  the  reaction  of  the 
wall  will  then  be,  say  g2c2,  inclined  below  the  normal  to  Ag  at 
the  angle  <p  (or  <p  if  <p  >  p),  and  the  reaction  of  the  plane  of 
rupture  will  lie  nearer  the  horizontal  than  its  normal  and  will 
make  the  angle  v  with  it. 

For  convenience,  lay  off  along  the  free  surface,  gi  =  12  =  23, 
etc.,  compute  the  weights  of  the  trial  wedges  of  rupture,  Agi, 


FIG.  1 6 


FIG.  17 


Ag2,  .  .  .  ,  and  lay  off,  to  the  scale  of  force,  ggi,  gg2,  .  .  . ,  equal 
to  these  weights.  (Any  weight,  as  that  of  Ag2,  is  found  by 
multiplying  ^  w.  g2  by  the  perpendicular  from  A  upon  #2.) 
The  next  step  is  to  draw  two  arcs  of  circles  with  A  and  g  as 
centers  and  a  common  radius  Ag  and  lay  off  the  angle  Ags  =  <p 
and  mark  the  intersections  of  Ai,  A2,  .  .  .  ,  with  the  arc  ag,  a\, 
02,  ....  Then  to  find  the  directions  of  the  reactions  of  the 
successive  trial  planes  of  rupture,  let  us  first  consider  one  of 
them  as  A2.  Draw  Aa  horizontal  and  let  a  =  angle  made 
with  the  vertical  Ag  by  the  normal  to  Ai  .'.  a  =  a  A  a2.  The 
direction  of  the  reaction  gs2  of  the  plane  A  2  must  make  an 
angle  <p  above  its  normal, 

.'.  Ags  +  sgsz  =  Ags2  =  <f>  +  a  =  Ags  +  a  Aa2', 
whence   sgsz  =  a  A  a2   or    chord    ss2  =  chord    aa2.     A    similar 
conclusion  holds  for  any  of  the  planes  Ai,  A2,  .  .  .  ,  so  that  to 
find  the  directions  of  their  reactions,  lay  off  ssi  =  aai,  ss2  =  aa2, 


36]  PASSIVE   THRUST  55 

ss3  =  aa3,  •  .  • ,  and  draw  gsi}  gs*,  gsa,  .  .  .  ,  to  give  the  directions 
sought. 

To  complete  the  triangle  of  forces  for  each  trial  wedge  of 
rupture,  draw  lines  parallel  to  the  wall  reaction,  through  gi, 
gz,  .  .  . ,  to  intersections  ci,  c2)  .  . . ,  with  gsi,  gs2,  . . .  ;  then  the 
least  of  the  intercepts  #3^3  is  approximately  the  passive  thrust 
and  A$  the  corresponding  plane  of  rupture.  This  follows,  be- 
cause if  the  thrust  is  any  greater  than  g3c3,  c3  moves  to  the  right 
and  the  new  gc3,  representing  the  reaction  of  A^,  lies  above  the 
old  position  and  thus  makes  a  greater  angle  than  <p  with  the 
normal  to  ^.3,  which  is  inconsistent  with  stability,  Art.  5. 

The  passive  thrust  (or  resistance)  is  much  larger  than  the 
active  thrust  and  incr eases  with  <p  and  <p.  To  keep  the  figure 
within  bounds  the  values  of  <pandV  were  taken  small.  When 
the  direction  of  the  thrust  on  the  vertical  plane  Ag  is  taken 
parallel  to  the  surface,  the  resulting  passive  thrust  equals  that 
on  a  vertical  plane  in  an  unlimited  mass  of  earth. 

For  passive  thrust,  the  plane  of  rupture  is  much  nearer  the 
horizontal  than  for  active  thrust.  The  surface  of  rupture  being 
longer,  is  doubtless  more  irregular  and  further  removed  from  a 
plane  than  for  the  case  of  active  thrust,  so  that  the  theory  is 
not  so  close  to  reality  as  in  the  latter  case.  The  construction 
of  Fig.  17  can  readily  be  extended  to  battered  walls  and  various 
slopes  and  surcharges  of  earth,  but  the  matter  is  not  of  sufficient 
practical  importance  to  enter  into  further. 

The  center  of  pressure  of  the  passive  thrust  is  on  Ag  at  a 
distance  (1/3)  Ag  above  A. 


CHAPTER  III 

NON-COHESIVE    EARTH.      ANALYTICAL   METHODS 

37.  Active  Thrust.  In  the  analytical  method  that  will  now 
be  given,  trial  wedges  of  rupture  will  be  assumed,  each  one  in 
turn  being  treated  as  a  true  one  until  the  application  of  the 
principle  of  Art.  5  enables  us  to  select  the  only  true  one  and 
to  find  the  corresponding  thrust. 

In  Fig.  18,  let  AF  represent  the  inner  face  of  the  wall  and 
FPQRC  the  earth  surface,  the  length  of  wall  considered  being 


FIG.  1 8 


i  foot,  perpendicular  to  the  plane  of  the  paper.  The  wall  AF 
will  be  supposed  to  resist  by  a  reaction  E,  inclined  at  the  angle 
<pf  (or  (p  when  <pf  >  <p)  to  the  normal  to  the  wall,  the  tendency 
of  the  earth  to  slide  down  some  plane  AC,  passing  through  A. 
Let  G  =  weight  of  earth  A  FPQRC,  contained  between  two 
planes  perpendicular  to  the  inner  face  of  the  wall  and  i  foot  apart, 
and  S  =  reaction  along  plane  AC,  inclined  at  the  angle  <p  to  the 
normal  to  AC,  the  component  of  S  parallel  to  AC,  acting  up 
along  AC.  Also,  let  AC  make  the  angle  7  with  the  horizontal 
A  H  and  the  angle  55  with  the  vertical.  Finally  let  a  =  angle 

56 


37,  38]  ACTIVE  THRUST  57 

made  by  AF  with  the  vertical.    The  values  of  certain  angles 
are  shown  on  the  figure. 

The  three  forces  E,  G  and  5  are  in  equilibrium  and  may  be 
represented  in  magnitude  and  direction  by  the  sides  W,  ON, 
NL  of  the  triangle  ONL,  E  and  5  acting  up  and  G  down.  Hence 
noting  that  ONL  =  y  —  <f>  and  that  NL  0  =  <f>  +  Z  +  a  +  <p', 
we  have,  by  the  law  of  sines, 

E      sin  ONL  sin  (  y—  <p) 

G  ~  sin  NLO  ~  sin  (<p  +  3  +  a  +  <p')     '      * 

As  AC  varies  its  position,  the  true  value  of  E  is  the  maximum 
obtained  by  the  construction  above  for  the  various  trial  wedges 
of  rupture;  for  if  A C  is  the  plane  of  rupture  corresponding  to 
this  greatest  trial  reaction  E,  if  E  is  taken  less  than  this,  the 
point  L  will  lie  nearer  0  and  NL  will  lie  nearer  the  vertical, 
hence  NL  or  S  will  make  an  angle  greater  than  <p  with  the  normal 
to  AC,  which,  Art.  5,  is  inconsistent  with  the  laws  of  stability 
of  a  granular  mass.  Hence  the  greatest  of  the  trial  thrusts  is 
the  least  thrust  consistent  with  equilibrium,  and  this  value  of 
E  is  the  magnitude  of  the  actual  active  thrust  exerted  against 
the  wall,  caused  by  the  impending  motion  of  the  wedge  AFPQRC 
down  the  plane  AC. 

38.  Let  us  now  consider  the  simpler  earth  profile  FRPQ  of 
Fig.  19  and  assume  that  the  true  plane  of  rupture  AC  intersects 
the  part  RP  of  the  profile.  If  FB  is  drawn  parallel  to  a  line 
from  A  to  R  to  meet  RP  produced  at  B,  then  the  areas  of  the 
two  triangles  A FR  and  ABR  are  equal  and  area  AFRC  is  equal 
to  that  of  the  triangle  ABC*  Hence  dropping  a  perpendicular 
A  T  from  A  upon  BC,  we  have  the  weight  of  the  wedge  AFRC,= 

G  =  -  w.  AT.BC,  where  w  —  weight  of  a  cubic  foot  of  earth 

in  pounds,  the  foot  being  the  unit  of  length.    In  Fig.  19,  draw 
the  line  CI  to  intersection  /  with  the  line  of  natural  slope  AD, 


*  In  Fig.  1 8,  the  point  B  is  found  on  the  prolongation  of  CR,  so  that  area 
AFPQRC  is  equal  to  area  triangle  ABC.  All  the  subsequent  conclusions  per- 
taining to  Fig.  19  apply  without  change  to  Fig.  18,  with  the  point  .5'thus 
found. 


58 


NON-COHESIVE  EARTH.      ANALYTICAL  METHODS 


making  the  angle  AC  I  =  (%  +  <p  +  <p  +  a).  Then,  noting 
that  (7  —  <p)  =  CAJ  —  DAJ  =  CAI,  we  can  replace  the  ratio 
of  the  sines  in  (5)  by  CI/AI,  so  that  (5)  can  be  written,  on 
substituting  the  above  value  of  G, 


E^-w.AT.BC.—. 


(6) 


FIG.  19 

On  drawing  BO  parallel  to  CI  to  intersection  0  with  AD, 
we  have  from  the  similar  triangles  BOD  and  CID, 

OI_          _         BO 

OD}  OD' 

whence  (6)  takes  the  form, 

'AT.BD.BO\     01.  ID 


A I 


(7) 


The  terms  in  (  )  may  be  reduced  to  a  simpler  form  on  noting 
that  AT.BD  represents  double  the  area  of  the  triangle  ABD 
and  can  be  replaced  by  AD.BN  =  AD. BO  cos  OBN,  where 
BN  is  perpendicular  to  AD\  whence, 

B0\2    01.  ID 

OD 


i  /B0\* 

E--w.ADco,OBN(—)    . 


(8) 


3SJ  ACTIVE   THRUST  59 

On  drawing  CH  perpendicular  to  AD,  we  note  that  AC  makes 
the  angle  55  with  a  vertical  at  C  and  that  CH  makes  the  angle 
<p  with  the  same  vertical,  the  sides  being  perpendicular  to  those 
of  the  angle  DAJ  =  <p.  Hence  since  ACT  was  laid  off  = 
(<S  +  <p  +  <p'  +  a),  it  follows  that  EC  I  =  NBO  =  (<?'  +  a). 
This  angle  being  constant,  BO  and  OD  are  constant  as  AC 
varies  in  position,  so  that  the  only  variable  part  of  (8)  is  the 
last  fraction.  For  brevity  write  A I  =  x,  AD  =  a,  AO  =  b, 
then  the  variable  term  can  be  written, 

01.  ID       (x-b)(a-x)  ab 

— — -  = =  a  +  b x 

AI  x  x 

which  is  a  maximum  for  x  =  A I  =  ^ab.  Substituting  this  value 
of  x  in  the  variable  term,  it  reduces  to, 

0-V^)2     YD* 

a  +  b  —  2  Vab  =  -  —  = , 

a  AD 

the  point  /  now  corresponding  to  the  -position  of  C  for  the  true 
plane  of  rupture  AC. 

On  substituting  this  maximum  value  for  the  variable  term 

BO      CI 

and  noting  from  the  similar  triangles,  BOD,  CID,  that =  — 

(8)  reduces  to, 

E  =  -  w.  CI 2  cos  HCI  =  -w.IL.CH  (o) 

2  2 

where  IL  has  been  laid  off  along  I A  equal  to  CI.  If  a  line  is 
drawn  from  L  to  C,  it  is  seen  that  the  active  thrust  E  against 
the  wall  in  pounds  is  exactly  represented  by  the  area  of  the  triangle 
IL  C  multiplied  by  w,  the  weight  per  cubic  foot  of  the  earth  in 
pounds* 


*  This  conclusion,  for  the  case  of  a  plane  earth  surface,  was  reached  by 
Rebhann  in  1871,  and  later  by  Weyrauch,  both  proceeding  along  independent 
lines,  different  from  the  above. 


60  NON-COHESIVE  EARTH.      ANALYTICAL  METHODS 

Since  the  angle  EC  I  =  (<pf  +  a),  the  formula  can  likewise 
be  written, 

E  =  -  w.  CI  cos  (p  +  a)    .     .     .     .      (10) 

All  the  preceding  results  refer  to  the  inner  face  of  the  wall  AF 
lying  to  the  left  of  the  vertical  through  A.  In  case  the  wall 
leans  toward  the  earth  or  AF  lies  to  the  right  of  the  vertical 
through  A,  all  the  preceding  formulas  hold  on  simply  changing 
a  to  (—a),  as  will  be  evident  on  reviewing  the  demonstration.* 

If  we  draw  Ay  parallel  to  CI  (or  to  BO),  since  CI  makes 
the  angle  (<p  +  <p  +  «)  with  the  vertical,  it  follows  that  Ay 
makes  the  same  angle  with  the  vertical,  hence  FAy  =  (p  +  <p. 
The  same  conclusion  holds  for  leaning  walls,  where  a  is  changed 
to  (—  a).*  It  is  tacitly  assumed  above  that  ^4Flies  above  the 
limiting  plane  and  that  the  conditions  outlined  in  Art.  28  or 
32  are  satisfied. 

39.  To  find  the  earth  thrust  E  against  the  plane  AF,  whether 
AF  lies  to  the  left  or  to  the  right  of  the  vertical  through  A  (in 
the  latter  case,  a  is  not  to  exceed  about  10°),  first  find  the  point 
By  so  that  area  triangle  ABR  =  area  triangle  AFR  and  draw 
the  line  Ay  making  the  angle  (<p  +  $>')  with  AF,  as  shown. 
Then  draw  BO  parallel  to  Ay  to  intersection  0  with  AD,  in- 
clined at  the  angle  p  to  the  horizontal.  Let  AD  intersect  RP, 
produced  if  necessary,  at  D.  Then,  x  =  A I  =  ^AD.AO  can: 
either  be  computed,  on  measuring  AD  and  AO  to  scale,  other- 
wise two  constructions,  given  by  geometry,  are  available  for 
locating  7. 

By  the  first,  a  semicircle  is  described  on  A  D  as  a  diameter 
and  at  the  point  O,  a  perpendicular  is  erected  to  AD,  meeting 
the  semicircle  in  M\  then  A I  is  laid  off  equal  to  the  chord  AM. 
By  the  second  construction,  a  semicircle  is  described  on  OD 
as  a  diameter,  a  tangent  to  it,  AG,  from  A  is  drawn,  limited  by 
the  radius  perpendicular  to  it,  and  finally  A I  is  laid  off  equal 
to  AG. 

*  See  Arts.  19  and  62  as  to  the  direction  of  the  thrust  to  assume  for  leaning 
walls. 


59,  40]  CONSTRUCTION  FOR  ACTIVE   THRUST  61 

The  point  /  having  been  located  by  calculation  or  by  either 
>f  the  constructions,  1C  is  drawn  parallel  to  Ay  to  intersection 
with  RPD'y  whence  AC  is  the  plane  of  rupture. 

On  measuring  the  length  CI,  also  the  length  of  the  perpen- 
licular  CH,  from  C  to  AD,  to  the  scale  of  distance,  the  earth 
;hrust  E  on  AF  is  given  by  X  w-  CI.CH;  otherwise  E  equals 
v  times  the  area  of  the  triangle  ICL,  where  IL  =  CI.  Where 
;he  dimensions  are  in  feet  and  w  in  pounds  per  cubic  foot,  the 
:hrust  E  is  given  in  pounds.  The  same  construction  applies 
;o  a  more  irregular  earth  contour,  as  in  Fig.  18,  as  soon  as  the 
Doint  B  is  found.  In  any  case,  if  A  C  is  not  found  to  intersect 
;he  free  surface  on  the  plane  assumed,  another  plane  may  be 
;ried,  etc.  For  a  curved  earth  contour,  the  general  method  of 
.he  preceding  chapter  can  be  used. 

40.  When  the  free  surface  is  plane  and  either  rises  or  falls 
ping  from  the  wall  or  is  horizontal,  the  construction  is  simpli- 
led  since  the  point  B  then  coincides  with  F,  which  generally 
alls  at  the  inner  upper  corner  of  the  wall.  It  may  lie  above  it, 
is  in  Fig.  19,  or  it  may  fall  below  it.  In  any  case,  AF  is  the 
;irea  pressed  along  the  inner  face  of  the  wall,  extended  if  necessary. 

Fig.  20  shows  the  construction  for  this  common  case,  AB 
:being  the  inner  face  of  the  wall.  The  letters  having  the  same 
ignificance  as  before,  the  thrust  E  =  w  X  area  triangle  CIL. 

Ex.  i.     Find  the  value  of  E  pertaining  to  Ex.  I,  Art.  35,  by  this  construc- 
ion. 

Ex.  2.     Find  the  value  of  E  for  some  of  the  examples  of  Art.  23. 

The  construction  is  a  good  one  except  when  angle  ICD  is 
mall,  when  the  point  C  cannot  be  located  very  accurately.  In 
that  case,  use  the  methods  of  Art.  21  or  perhaps  the  construc- 
tion, Fig.  22.  Both  the  graphical  methods  fail  when  the  in- 
clination of  BD  to  the  horizontal  is  large,  on  account  of  the 
ntersection  D  being  too  remote. 

In  that  case,  formulas  will  have  to  be  resorted  to.  In  Fig. 
20,  as  BD  approaches  the  natural  slope  or  approaches  parallel- 
sm  to  AD,  the  distance  AD  increases  indefinitely  and  so  does 
A I  =  ^JAD.AO,  so  that  the  triangle  CIL  approaches  in 


62 


NON-COHESIVE  EARTH.      ANALYTICAL  METHODS 


size  the  similarly  constructed  triangle  with  /  anywhere 
on  AD  and  1C  II  Ay,  meeting  a  line  through  B  parallel  to  AD 
at  C.  The  area  of  the  triangle  CIL  corresponding,  multiplied 
by  w  gives  the  thrust  E  when  BCD  is  at  the  natural  slope. 
Also  regarding  now  C  as  in  true  position,  CI  being  finite,  since 


FIG.  21. 


AC  and  A I  approach  infinity  with  AD,  it  follows  that  the 
plane  of  rupture  AC  approaches  indefinitely  AD  or  the  natural 
slope  as  AD  indefinitely  increases  or  as  BD  approaches  in- 
definitely the  natural  slope. 

As  a  special  case,  take  AB  vertical  and  let  C  coincide  with 
B,  Fig.  21,  the  earth  surface,  CP  being  at  the  natural  slope. 
Then  if  E  makes  the  angle  *p  —  <?  with  the  normal  to  AB, 
/  ACI  =  <p  +  (p  —  2(p;  whence  /  A  1C  =  (go  —  2(p)  +  <p  = 
90  —  (p  =  Z  I  AC  .'.  CI  =  AB  =  h.  .'.  since  a  =  o,  by 

do), 


E  =  -  w  CI  cos  <p  =  -w 

2  2 


cos 


the  well  known  formula  for  the  thrust  on  a  vertical  plane  in  an 
unlimited  mass  of  earth  with  the  surface  sloping  at  the  angle 
of  repose.  This  formula  gives  a  large  excess  to  the  true  thrust 
in  any  practical  case,  since  an  infinite  plane  of  rupture  is  never 
realized  for  any  possible  back  filling.  See  Art.  43,  Fig.  23. 

Ex.  i.     Find  the  thrust  against  the  vertical  plane  BH,  Fig.  1 6  in  Ex.  5, 
Art.  35. 

Ex.  2.     Check  some  of  the  values  of  K  in  Art.  31. 

41.  Several   interesting   conclusions    follow   from   the   con- 
struction, Fig.  20.     Observing  that  if  a  line  is  drawn  from  the 


1] 


CONSTRUCTION   FOR  ACTIVE   THRUST 


63 


enter  of  the  semicircle  parallel  to  1C,  it  meets  BD  at  its  mid- 
>oint,  it  is  seen  that  C  is  to  the  left  of  this  mid-point,  but  ap- 
>roaches  it  as  BD  approaches  BO,  since  when  OD  is  very  small, 
he  point  /  and  the  center  of  the  semicircle  are  nearly  coincident. 

At  the  limit,  when  BD  falls  upon  the  fixed  line  BO,  C  is  ex- 
,ctly  at  the  mid-point  of  BD  =  BO. 

When  BD  falls  below  BO,  D  lies  below  0,  but  the  semi- 
ircle  is  described  on  DO  as  diameter,  since  A I  =  *\AD.AO, 
,s  before;  but  1C,  drawn  parallel  to  BO,  now  meets  BD  to  the 
ight  of  its  mid-point. 

In  Fig.   22  is  given  another  construction  for  locating  the 


B         C 


m,  say, 


FIG.  22a 

>oints  C  and  /.     Extend  the  line   DB  to  meet   the  line  Ay 
t  y\   then  from  similar  triangles, 

AI  =  AO  =  AD 
yC  "  yB  "  yD 

n  substituting,  A I  =  m.  yC,  etc.,  in  the  relation  previously 
stablished, 

AP  =  AO.AD 

nd  striking  out  m2,  we  have, 

yC2  =  yB.yD. 

On  describing  a  semi-circle  on  BD  as  diameter,  drawing 
>z  (limited  by  a  radius  perpendicular  to  it)  tangent  to  the  semi- 
ircle,  then  from  the  last  equation,  yC  is  to  be  laid  off  equal  to 
>z  to  locate  the  point  C* 


*See  this  construction  in  Prelini's  "Earth  Slopes,  Retaining  Walls  and 
}ams,"  where  Rebhann's  neat  analysis  is  given. 


64  NON-COHESIVE    EARTH.       ANALYTICAL    METHODS 

The  plane  of  rupture  AC  is  thus  determined  more  direct 
than  by  the  preceding  construction.  On  drawing  CI  paral] 
to  Ay  to  intersection  /  with  AD,  the  thrust  E  =  >£  w.  CI  ] 
perpendicular  from  C  on  AD,  as  given  by  eq.  (9). 

It  will  now  be  proved,  for  a  perfectly  smooth  vertical  we 
(<pf  =  o,  a  =  o),  with  earth  surface  horizontal,  that  the  plane 
rupture  bisects  the  angle  between  the  vertical  and  the  line  of  natur 
slope.     In  Fig.   22  (a),   conceive   AB   to  be  vertical  and   B. 
horizontal;   then,   since   <p  =  o,   ABO  =  <p,   and   since   BD 
parallel  to  AJ,  BDA  =  <p.    Hence  the  two  right  triangles,  AB( 
ABD,  are  similar, 


t  follows  that  AB  —  AI,  whence  the  two  right  triangles  BA 
CAI  are  equal 

.*.    /  BAG  =  Z  CAI  =-j  (90°  -  <?), 

which  was  to  be  proved. 

The  horizontal  thrust  against  the  vertical  wall  AB  is, 

E  =  —w.CP  =  —w.BC2=—w.AB2tan2(4S°-~}  •  .  (xi 

222  \     •  2' 

This  is  the  well  known  expression  for  the  horizontal  thru: 
against  a  vertical  plane,  in  an  unlimited  mass  of  earth  with 
horizontal  surface.* 

42.  Historical  Note.  The  sliding-  wedge  theory  of  eart 
pressure  was  first  put  upon  a  scientific  basis  by  Coulomb  (1773 
He  assumed  <?'  =  o,  whereas  Poncelet  (1840)  took  <?'  at  its  tru 
value  and  developed  in  an  elegant  manner  a  graphical  metho 
equivalent  in  its  results  to  the  above.  Since  Poncelet's  time, 
host  of  authors  have  developed  the  sliding-w^dge  theory  alon 
more  or  less  original  lines.  In  particular,  Reohann  in  187: 
deduced  formula  (9)  of  Art.  38  with  the  corresponding  cor 

*  In  the  author's  "Practical  Designing  of  Retaining  Walls"  (Van  No 
trand's  Science  Series),  pp.  90-93,  is  given  a  very  simple  diagram  for  findir 
the  Rankine  thrust  on  a  vertical  plane,  the  surface  sloping  at  any  angle,  an 
the  plane  of  rupture  (limiting  plane)  corresponding,  the  analysis  being  simile 
to  that  which  precedes.  It  is  omitted  here,  because  the  same  subjects  will  t 
fully  considered,  from  another  standpoint,  further  on. 


66 


NON-COHESIVE    EARTH.       ANALYTICAL  METHODS 


applied  to  a  vertical  wall,  the  earth  surface  sloping  at  the  angle 
of  repose  from  B  to  R,  then  horizontal.  Assuming  <p  =  /  = 
33°  41'  and  calling  the  vertical  height  of  wall  AB  =  h  and  the 
height  of  surcharge  BC  =  h',  Fig.  23,  the  planes  of  rupture 
shown  by  the  broken  lines  and  the  thrusts  were  found  for  a 


FIG.  23 

constant  hf  =  10'  and  for  h  =  i,  2,  .  .  .  ,  20  ft.,  in  turn.  The 
difference  between  the  successive  thrusts  gave  the  pressures 
on  each  successive  foot  (shown  by  the  arrows  in  magnitude 
only).  Assuming  these  to  act  at  the  center  of  the  divisions, 
the  centers  of  pressure,  for  h  varying  from  5  to  20  ft.,  were  found 
by  taking  moments  about  convenient  points.  Letting  c  = 


r2.  4";  CENTER   OF   PRESSURE  65 

struction  and  later,  Weyrauch,  by  an  independent  analysis. 
reached  similar  conclusions,  the  direction  of  the  thrust  on  the 
vrall,  however,  not  being  assumed,  but  forming  part  of  the 
problem. 

Rankine  in  1856  from  considerations  of  certain  properties 
of  stress  (really  affecting  an  infinitesimal  wedge  of  a  granular 
material),  put  the  theory  of  earth  pressure  on  a  vertical  plane 
in  an  unlimited  mass  of  earth,  on  a  thorough  and  exact  scientific 
basis.  It  is  only  in  his  proposed  application  of  this  theory  to 
rough  retaining  walls  that  criticism  can  be  made.  Boussinesq 
attempted  to  complete  the  Rankine  theory  by  considering  the 
influence  of  the  wall  friction  in  modifying  the  Rankine  thrust. 
His  analysis  only  leads,  in  a  simple  case,  to  two  limits  to  the 
thrust,  whose  most  probable  value  is  the  smaller  of  these  limits 
augmented  by  9/22  of  their  difference.  The  wrork  of  Resal 
(1903)  has  already  been  alluded  to  in  Art.  33. 

The  author  has  been  careful  to  point  out  in  Arts.  26-30  where 
Rankine's  theory  applied  to  retaining  walls  gives  exact  results 
and  where  inexact  results.  In  the  latter  case,  the  sliding-wedge 
theory  is  resorted  to,  for  though  it  is  admittedly  approximative 
for  such  cases,  it  is  easily  understood  and  answers  the  needs 
^of  practice  in  ordinary  cases.  The  sliding-wedge  theory  gives 
exact  results  for  the  case  of  the  unlimited  mass  of  earth  and, 
in-fact,  leads  to  the  Rankine  formula,  Art.  48. 

43.  Center  of  Pressure.  Referring  to  Fig.  20,  it  is  seen 
that,  for  the  same  surface  slope  and  inclination  of  AB,  if  the 
vertical  height  h  of  AB  varies,  then  CI  (always  parallel  to  Ay) 
will  vary  directly  with  AB  or  with  //.  Hence  the  thrust  E, 
which,  by  (9),  varies  as  C/2,  will  vary  as  h-  and  car.  ^  repre- 
sented by  an  expression  of  the  form,  E  =  K  w  7/2,  where  A'  is  a 
constant  for  given  inclinations  of  the  wall  and  free  surface.  It 
follows,  as  in  the  case  of  water  pressure,  that  the  center  of 
pressure  on  AB  will  be  at  one- third  its  height  above  A.  Com- 
pare with  Art.  24. 

For  the  case  of  the  surcharged  wall  of  Fig.  19,  the  center  of 
pressure  is  higher  than  the  third  point  going  from  A  to  F.  To 
determine  it,  a  construction  similar  to  that  of  Fig.  19  was 


4: 


atio  of  distance  from  foot  of  wall  considered  to  center  of  press- 
ure to  height  of  wall,  the  results  are  as  follows: 


A' 

k' 

h, 

V 

T 

C 

k 

c 

k 

c 

k 

c 

00 

0-333 

1-50 

0.356 

I.OO 

0.364 

.... 

1-25 

0.360 

0-75 

0.364 

.... 



2.OO 

0-353 

I.  II 

0.362 

0.50 

0.364 

O 

0-333 

SURFACE    SLOPING   UNIFORMLY.      FORMULAS 


67 


The  thrusts  were  assumed  to  make  the  angle  $  =  <p  with 
le  horizontal,  though  here,  the  arrows  giving  their  magnitudes 
ivere  laid  off  horizontally  to  show  the  variation  better.  For 
ther  values  of  <p  and  $>',  the  values  of  c  may  change  somewhat, 
ut  they  are  probably  near  enough  to  use  in  practice  for  usual 
alues  of  <p. 

44.  Surface  Sloping  Uniformly.  Formulas  for  E.  The  for- 
mula for  E  is  easily  derived  from  Fig.  24,  where  the  letters 
lave  the  same  meaning  as  in  Fig.  20. 

Let,  <p  =  angle  of  repose  of  earth  =  JAD, 

<?'  =  angle  of  friction  of  earth  on  wall  AB, 
i  =  angle  made  by  free  surface  with  horizontal, 
a  =  angle  made  by  inner  face  of  wall  AB  with  the 

vertical, 
h  =  vertical  height  of  wall  —  AB  cos  a. 

As  in  Fig.  20,  we  have  ABO  —  BAy  =  <?-{-<?'  and  from 
ig.  19,  NBO  =  *>'  +  a  /.  NOB  =  90°  -  (/  +  a).     We  also 
iave,      * 
ADB  =  <p  -  »,  ABD  =  90°  -  (a  -  0,  BAO  =  90°  +  a  -  ?. 

In  (8),  Art.  38,  on  substituting  for       '      ,  its  maximum 


alue, 


a- 


AD 


,  we  have  the  active  thrust  given  in  the  form, 


rhere,  a  =  AD,  b  =  AO. 


68  NON-COHESIVE    EARTH.       ANALYTICAL    METHODS 

Let,  n  =  \!  —  =  \l  *——  and  write  the  bracketed  term  as 
™  a       ™  AD 

follows : 


a  — 


a-l 


b  IT 

i i  +  \  -        i  +  n 

a  *  a 


FIG.  24 

To  find  n,  from  the  triangles  AOB  and  ABD,  by  the  law 
of  sines, 

A  O  _  sin  ((p  +  <p')    A  B      sin  (<p  —  i) 
AB     '  cos  O'  +  a)'  AD  ~  cos  (a  -  i) 

Multiply  these  two  equations  together  and  extract  the  square 
root: 

\AO__       I  sin  (<p  +  <ff)  sin  (<P  -  i) 
'^  AD       X  cos   (9'  +  a)    cos   (a  -  i) 

From  the  triangle  BOA  ,  by  the  law  of  sines, 
BO       cos   (<p  —  a) 

COS  (<?'  +  a)* 


Substituting  in  (13),  -  for  the  bracketed  term  and  the  value 


n 


of  BO  obtained  from  the  last  equation,  also  writing  cos  OBN  = 
cos  (tp'  -f-  a)  we  have, 

.—  /  \  n  i     ^-^ 

iTcos  (<f>  — 
E  =  - 


n+  f          cos 


44, 45]  SURFACE   SLOPING.  UNIFORMLY.      FORMULAS  69 

or  since,  AB  = , 


cos  a 


I  [~~  COS  (<p  —    a)  — i 2  w  h* 

=  7L(l+")   cosa  J   COS   (</  +  a)    '       '       '  .  ^ 

On  putting, 

i  r  cw(^-a)  -i2  i 

^TUi  +  tt)  cojtt-1    ^TT+"a)    '       ' 

having  the  value  given  by  (14),  we  can  write  (16), 

E  =  Kwh*    '  .     .     ...     .    (18) 

The  preceding  results  all  refer  to  walls  with  the  inner  faces 
mattered  or  those  leaning  away  from  the  earth. 

For  overhanging  walls  or  those  leaning  toward  the  earth,  a 
)eing  the  angle  made  by  the  inner  face  with  the  vertical,  for- 
mula (18)  again  applies  provided  we  put, 


f  sin  (<p  +  <f>')  sin  (<P  —  i)  ,    . 

:    \C07V-  a)  cos  («  +  /)       '      '      <; 


and, 

~  cos 


a)  Y  _  !__  ,    v 

S  a-*      COS     <p'—  a 


+  n)   COS  a-*      COS   (<p'—  a) 

The  last  two  formulas  are  easily  seen  to  be  true  by  drawing 
figure  and  noting  that  the  new  angles  involving  a  can  be 
ound  from  the  old  by  simply  changing  a  to  (—  a).     To  avoid 
mistake,  in  all  the  formulas  (14)  —  (20),  a  is  taken  positive. 

//  is  important  to  note  that  the  formulas  or  graphical  methods 
)f  this  chapter  are  only  applicable  to  the  cases  cited  in  Art.  30 
>r,  usually,  when  AB  lies  above  the  limiting  plane,  and  always 
or  walls  leaning  towards  the  earth  filling,  when  a  does  not 
xceed  about  10°.  Also  see  Art.  62. 

45.  The  case  where  the  free  earth  surface  slopes  downward 
o  the  rear  is  sometimes  though  rarely  met  with  in  practice. 
All  the  preceding  formulas  apply  directly  on  simply  changing 
i}  to  (—  i),  as  is  easily  seen  on  Drawing  a  figure. 


70  NON-COHESIVE    EARTH.       ANALYTICAL    METHODS 

NOTE.  The  formulas  above,  for  all  the  cases  mentioned,  are  very  long 
and  tedious  in  the  applications,  except  for  the  special  cases  to  be  given  in 
the  next  article.  For  all  other  cases,  it  wil)  shorten  the  labor  to  measure 

AO  and  AD  on  a  drawing  to  a  large  scale  and  compute  n  =^l  -— ,  by  using 

these  values;  then  if  AO  and  AD  have  been  scaled  approximately  to  four 
significant  figures,  K  can  be  computed  accurately  to  three,  which  generally 
answers  the  needs  of  practice. 

46.  Earth  Level  at  Top;  back  of  wall  vertical.  Earth  Sloping 
at  angle  of  Repose.  For  a  horizontal  earth  surface,  back 
of  wall  vertical  and  $  =  <f>,  (16)  takes  a  simple  form.  On 
substituting  a  =  o,  /  =  <p,  i  =  o  in  (14),  we  have, 


.  I  sin  2 

»->/- 


&  sin  <p        .         / — 
-  =  sin  <p  V2 

COS  (p 

whence  by  (16) 

1  w  h*  cos<p 

2  (i  +  sin  <?  V2  )2 

If  in  place  of  taking  y  =  <?,  we  assume  <?'  =  o,  which  may 
either  refer  to  a  perfectly  smooth  wall  or  to  the  horizontal  thrust 
on  a  vertical  plane  in  an  indefinitely  great  mass  of  earth  with 
level  surface,  the  formula  for  E  is  still  further  simplified.  Now 
we  have,  a  =  o,  i  =  o,  <p'  =  o,  n  =  sin  (p. 

i  —  sin  <p  w  hz       i  /  <p  \ 

i  +  sin  <p    2         2  ^  2  ' 

The  equality  of  the  two  trigonometric  terms  can  be  verified  by 
putting  (90  —  <p)  for  x  in  the  known  formula, 


tari< 


X         I  —  COS  X 


2          I  +  COS  X 


Eq.  (22)  was  deduced  in  another  way  in  Art.  41,  where  it  was 
further  shown  that,  for  this  case,  the  plane  of  rupture  bisected 
the  angle  between  the  vertical  and  the  line  of  natural  slope. 

Earth  surface  at  angle  of  repose.    For  this  case,  i  =  v  and 
for  a  vertical  wall  a  =  o,  n  =  o;  whence  (16)  reduces  to, 


46-48]  RANKINE'S  FORMULA  71 

i         cos2 
E=—wh2 

2 

and  E  makes  the  angle  <p'  with  the  horizontal.  If  we  take 
<P'  =  <?,  the  thrust  E  acts  parallel  to  the  sloping  surface  and  we 
derive  again  the  formula  of  Art.  40, 

E  =  —  w  h2  cos  <f> (23) 

47.  Pressure  of  Fluids.    As  <?,  <?'  and  i  approach  zero,  the 
granular  mass  approaches  the  state  of  a  perfect  fluid,  so  that 
at  the  limit,  when  <p  =  <?'  =  i  =  o,  we  have  from  (14),  n  =  o 
and  from  (15), 

E  =  -  w.  AB2  cos  a  =  -  w.  AB     ,     .     .!  (24) 

2  2 

or  the  surface  pressed,  multiplied  by  w  times  the  vertical  dis- 
tance from  the  level  surface  to  the  center  of  gravity  of  the  surface, 
a  well  known  formula  for  the  normal  pressure  of  a  perfect  fluid 
weighing  w  Ib.  per  cu.  ft. 

It  is  interesting  to  note  that,  when  a  =  <p'  =  i  =  o,  the  plane 
of  rupture  approaches  an  inclination  of  45°  to  the  horizontal  as 
(p  approaches  zero,  Art.  41. 

48.  Rankine's  formula  for  the  thrust,  acting  parallel  to  the 
sloping  surface,  on  a  vertical  plane  in  a  granular  mass  of  in- 
definite extent  and  not  subjected  to  any  external  force  but  its 
own  weight.    In  Art.  26,  it  was  shown  that  in  such  a  mass  the 
thrust  on  the  vertical  plane  could  only  act  parallel  to  the  surface; 
hence  putting  a  =  o,  <f>  =  i  in  (14)  and  (16),  we  have, 


n  =  -     -  A/  sin  (<?  +  i)  sin  (<?  —  i) 
cos  i 

—  V  sin2  (p  cosz  i  —  cos2  <f>  sin2  i 


cos 
i 


cos2  i  —  cos2  <p. 
cos  i 


72  NON-COHESIVE    EARTH.       ANALYTICAL    METHODS 

wh2    t  cos  <f>  \ 2 


2  COS  i   ^1  • 

•wh2 

-f-  n' 
cos2  (p  cos  i 

2     (cos  i 

1      9  • 

\2 

(ft    1 

Substituting/ 

cos2  (p  =  (cos  i  -\-  *\cos2  i  —  cos'2  <p) 

X  (cos  i  —  ^cos2  i  —  cos2  <f>) 
and  striking  out  the  common  factor,  we  obtain, 


wh2  cos  i  —  -\  cos2  i  —  cos2  <p     .      .    (25) 

E  —  -  cos  i 


, 
cos  i  +  \  cos2  i  —  cos2 


Rankine's  well  known  formula. 

This  formula  will  be  proved  in  Art.  58  from  certain  general 
considerations  affecting  stress,  not  assuming  a  plane  of  rupture 
extending  to  the  surface.  It  is  recognized  as  exact  by  all 
authorities. 

The  assumption  of  a  plane  surface  of  rupture  for  the  un- 
limited mass  is  then  justified,  for  it  leads  to  (25),  an  exact 
formula.  The  graphical  construction  of  Art.  26,  Fig.  10,  .is  thus 
correct,  but  that  of  Art.  21,  Fig.  9,  or  of  Art.  32,  Fig.  n,  must 
be  regarded  as  approximate.  Similarly,  the  construction,  Art. 
40,  Fig.  20,  will  give  the  exact  thrust  on  a  vertical  plane  in  an 
unlimited  mass,  when  AB  is  taken  vertical  and  </  is  replaced 
by  i.  The  construction  for  all  other  cases  is  approximate. 

It  may  be  remarked  that  the  two  special  formulas,  (22)  and 
(23),  may  both  be  deduced  from  (25)  by  putting  i  =  o,  i  =  <f>, 
respectively. 

49.  Unit  Pressure.     On  putting, 


1  cos  i  —  A/  cos2  i  —  cos2  <p 
K  =  -cos  i-  , 

2  cos  i  +  -\  cos2  i  —  cos2  (p 

we  can  write  (25)  in  the  form, 

E  =  K  T£*  h2. 


49,  50]  UNIT  PRESSURE  73 

Regarding  h  as  the  variable  depth  from  the  surface  down  to  a 
point  on  the  vertical  plane  considered,  let  (h  +  A  h)  represent 
a  depth  greater  by  A/?,  corresponding  to  a  thrust  (E  +  A  E), 
where  A  E  represents  the  increase  in  the  thrust. 

.'.  E  +  A  E  =  K  w  (h  +  A  h}* 

On  subtracting  the  first  equation  from  the  second  and  divid- 
ing by  A  h, 

A£ 

-  =  Kw(2h  +  A  A). 
A// 

Now  regarding  A  h,  which  equals  A  h  X  i  ,  as  the  area  on 
which  A  E  acts,  the  left  member  gives  the  average  pressure 
per  sq.  ft.  over  the  area  A/z  X  i  and  the  limit  of  this,  as  A// 


tends  indefinitely  towards  zero,  is  defined  toTfae,  the  intensity 
of  the  pressure  at  the  depth  x.  Hence,  in  the  notation  of  the 
calculus, 

dE 

Intensity  at  depth  //  =  --  =  2  K  w  h      .      .    (26) 
dh 

The  intensity  of  pressure  per  sq.  ft.  thus  varies  directly  as  the 
depth  and  can  be  represented  by  the  ordinates  of  a  triangle  with 
one  vertex  at  the  surface  on  a  vertical  side. 

Putting  p  =  wh  =  vertical  pressure  in  Ib.  per  sq.  ft.  at 
depth  h,  we  also  have,  intensity  =  2  K  p. 

Similarly  putting  /  =  AB  in  (15),  referring  say,  to  Fig.  20 
or  Fig.  22  and  regarding  /  as  variable,  the  intensity  of  pressure 

dE      - 
per  square  unit  of  the  face  AB,  at  the  point  /,  is  —  -. 

.  50.  Center  of  Pressure  for  Uniformly  Sloping  Surface.  The 
normal  component  of  the  thrust  on  the  plane  AB,  whether  it 
leans  forward  or  backward,  is  found  by  multiplying  (15)  by 
cos  v  and  it  can  be  expressed  by  the  simple  relation, 


c  being  a  constant  for  given  (p,  <?',  a  and  i.     Whence  the  thrust 


74  NON-COHESIVE    EARTH.       ANALYTICAL   METHODS 

on  the  area  dl  X  i,  at  the  variable  distance  /  from  B,  measured 
along  BA ,  dl  being  small,  is  nearly, 

dEi  =  2  cl .  dl, 

and  its  moment  about  the  point  B,  Fig.  20,  is,  2  cl2  .dl\  hence 
the  distance  from  B,  measured  along  BA,  to  the  center  oi 
pressure,  is  equal  to  the  limit  of  the  sum  of  such  moments3 
divided  by  the  total  pressure  or, 

2d*.dl       2 


hence  the  center  of  pressure  on  AB  is  at  one- third  its  height 
above  A. 

51.  Surcharge   Load,  Uniformly  Distributed.     In  Fig.  25  is 
shown  a  wall  A  BCD  backed  by  earth  ABP  with  a  level  surface 


~E>       l\\ 


FIG.  25 


A  J 

FIG.  26 


BP,  on  which  a  uniformly  distributed  load  rests.  This  load 
may  consist  of  buildings  or  locomotives  resiib.g  on  railroad 
tracks  parallel  to  the  wall. 

Let  W  =  weight  of  this  load  in   Ib.  per  sq.  ft.  of  NP  and 
h'  =  height  of  earth,  weighing  u>  Ib.  per  cu.  ft.  of  the 

same  weight  as  the  load. 

.*.  w  hf  =  W,  so  that  h'  =  MN  can  be  computed. 
It  represents  the  height  of  earth  that  will  be  supposed  to  replace 
the  load,  since  its  weight  is  the  same. 


51]  SURCHARGE  LOAD,   UNIFORMLY  DISTRIBUTED  75 

In  the  case  of  such  surcharges,  it  is  a  common  and  commend- 
able practice  to  assume  the  thrust  on  the  vertical  plane  AN  as 
acting  horizontally  and  to  compute  its  amount  by  aid  of  (22) 
by  taking  the  difference  of  the  thrusts  on  AM  and  MN,  or  to 
proceed  by  first  computing  the  unit  thrusts  at  A  and  N  by  use 
of  (26).  By  the  last  method,  the  unit  thrust  in  the  filling  at 
N,  which  is  caused  by  the  load,  p  =  wh',  is  2  K  wh'.  Lay  off, 
to  scale,  NO  equal  to  this  unit  thrust.  Similarly,  lay  off  AJ  = 
2  Kw.  AM.  The  straight  line  JO  should  pass  through  M. 
The  total  thrust  on  A  N  is  the  mean  of  the  thrusts  at  A  and  N 
multiplied  by  the  area  AN  X  i. 

By  the  first  method,  letting  AN  =  h,  NM  =  h',  the  total 
horizontal  thrust  on  AN  is, 

E  =  Kw[(h  +  hf)*-h">]  =  Kwh(h  +  2k')     .    (27) 
In  this,  as  well  as  in  the  preceding  method, 


Differentiating  the  expression  for  £, 

dE  =  2  K  w  (h  +  V)  dh, 

and  the  distance  of  the  center  of  pressure  /  on  .4  AT",  from  N 
downwards  is, 

J"  h 
0   (h  +  h  )  h  d  h        2   2  h  +  3  h 

h(h  +  2  h')  3     h  +  2  h'  ' 

and  from  A  upwards, 

AI  =  k  -  - —  = —t  .  — 

$h  +  6h         h  +  2h        3 


(28) 


This  thrust  E  acting  horizontally  to  the  left  at  7,  is  com- 
bined with  the  weight  of  filling  ABN  and  the  weight  of  wall, 
to  find  the  resultant  on  its  base.  Even  if  the  surcharge  extends 


76      NON-COHESIVE  EARTH.   ANALYTICAL  METHODS 

to  B,  it  is  wise  to  ignore  the  portion  over  BN  since  the  surcharge 
load  is  never  accurately  known,  even  for  warehouses,  and  train 
loads  are  continually  increasing.  On  that  account  this  solution 
is  advised  in  place  of  the  wall-friction  method.  There  are 
many  applications  of  it  in  the  chapter  on  concrete  walls. 

For  future  reference,  it  may  be  well  to  note  that,  E  X  A I  = 

—  K  w  h2  (/z  +  3  hr) ;  also  the  plane  of  rupture  A  F  bisects  the 
3 

angle  between  AN  and  the  natural  slope,  Art.  41.  Where 
the  rilling  is  of  the  type  shown  in  Fig.  n,  Art.  32,  with  a  load 
on  the  level  portion,  it  can  be  reduced  to  an  equivalent  weight 
of  earth  and  the  thrust  found  by  the  construction  of  Art.  32. 
Its  point  of  application  on  the  wall  is  unknown,  but  it  may  be 
taken  at  0.4  the  height  of  AB  ordinarily. 

52.  Dock  Walls.  If  the  water  rises  to  certain  heights  above 
the  base,  Fig.  25  can  be  taken  to  represent  a  sea  wall  or  a  dock 
wall.  Let  A N  in  Fig.  26  represent  the  vertical  plane  through 
A  of  the  preceding  figure  and  suppose  the  tide  at  high  water 
reaches  the  level  MP  and  at  low  water  a  lower  level.  If  the 
water,  at  high  tide,  reaches  the  same  height  on  the  front  and 
back  faces  of  the  retaining  wall,  the  water  pressures  on  those 
faces  will  balance,  but  if  there  is  a  probable  lag  of  the  tide  back 
of  the  wall,  its  dimensions  may  have  to  be  arbitrarily  increased. 
In  Fig.  26,  the  earth  will  be  regarded  as  dry  above  the  level 
MP  and  saturated  below.  Let  hi  =  NM,  h2  =  ML  and  h-3  = 
LA  and  call  the  horizontal  earth  thrusts  on  NM,  ML  and  LA, 
Ei,  Ez,  £3,  respectively.  Also  let  (wi,  <&),  (w2,  w),  (w>3,  ^3), 
represent  the  weights  per  cu.  ft.  and  the  angles  of  repose,  of  the 
earth  above  MP,  between  LO  and  MP  4rd  below  LO 
respectively. 

As  before,  h'  wi  —  W  =  load  on  NQ  in  pounds  per  square 
foot  of  NQ  and  from  (27), 

Ei  =  -  tan*  (4S°  ~  ~)  Mi2  +  2  hiW\ 

2  2  ' 

and  from  (28),  EI  acts  above  MP  the  distance, 


52]  DOCK   WALLS  77 


The  uniform  load  per  sq.  ft.,  on  the  plane  MP  is  now,  W  = 
W  +  wihi  =  w2A",  where  h"  is  the  height  of  an  equivalent  load 
of  earth  whose  weight  in  Ib.  per  cu.  ft.  is  wz  or  that  of  the  middle 
stratum. 

By  (27)  and  (28), 

1  /  ^2  \ 

EZ  —  —  tan1  (45° )  [wvhz2  +  2  h*  W'\ 

2  V  2  / 

and  it  acts  above  L,  the  distance, 

h"       v  A, 

Similarly,  we  replace  the  load  per  sq.  ft.  on  LO,  W'  +  wzh-2  = 
W" ,  by  an  equal  weight  of  earth  of  weight  in  Ib.  per  cu.  ft., 
w>3  and  height  ti",  so  that,  ti"wz  =  W" ,  from  which  /?'"  can 
be  computed  .".  by  (27)  and  (28), 

1  /      o       ^; 

2  ^  2 

and  £3  acts  above  A  the  distance, 


Having  found  by  these  formulas,  Ei,  £2,  £3,  in  position  and  mag- 
nitude, the  position  of  their  resultant  can  be  found  by  taking 
moments  about  A .  This  resultant  must  now  be  combined  with 
the  weight  of  earth  and  wall  to  the  left  of  the  plane  AN,  to  get 
the  final  resultant  on  the  base.  The  weight  of  the  submerged 
filling,  is  the  weight  in  water,  as  found  by  aid  of  the  table  in 
Art.  9.  The  weight  of  masonry  above  high  water  will  be  that 
in  air;  below  high  water,  the  weight  in  Ib.  per  cu.  ft.  must  be 
diminished  by  64  for  salt  water.  The  position  of  the  vertical 
through  the  center  of  gravity  of  the  combined  weights  of  sub- 
merged and  unsubmerged  earth  (to  left  of  AN)  and  masonry, 


78 


NON-COHESIVE    EARTH.       ANALYTICAL    METHODS 


can  be  found  by  taking  moments  about  A  or  any  convenient 
point. 

It  is  very  desirable  to  secure  rilling  with  as  large  an  angle 
of  repose  in  water  as  possible.  Thus  river  mud  is  undesirable, 
<p  varying  perhaps  from  o°  to  15°;  for  sand  and  clay,  dry,  say, 
<p  =  36°  53',  wet,  <p  =  18°  26';  whereas  rip-rap  or  cobblestones 
will  have  <p  =  42°  to  45°,  dry  or  submerged.  Hence  the  latter 
is  the  most  desirable  rilling  of  all. 


FIG.  27 

To  illustrate  with  a  numerical  example,  let  rip-rap  be  placed  below  JO 
extended  to  the  wall,  JO  making  an  angle  of  45°  with  the  horizontal  and 
suppose  a  sand  and  clay  mixture  above  OJ  to  extend  to  the  level  NQ.  E2 
will  be  computed  for  the  saturated  earth.  All  lengths  given  are  in  feet,  weights 
in  pounds.  Assume;  hi  =  5*  wi  =  IO°»  <Pi  —  33°  41',  for  dry  earth;  hz  =  6, 
W2  =  65,  <f>2  =  1  8°  30',  for  saturated  earth;  hs  =  10,  ws  =  65,  <p3  =  45°, 
for  rip-rap  in  water.  Then  since  the  plane  of  rupture  bisects  the  angle  be- 
tween the  vertical  and  the  line  of  natural  slope,  Art.  41,  the  planes  of  rupture 
make  angles  with  the  vertical  of  22°  30'  for  rip-rap,  35°  45'  for  saturated 
earth  and  28°  10'  for  dry  earth.  AO,  OP,  PQ,  represent  these  successive 
planes  of  rupture,  so  that  ANQPOA  is  the  mass  that  would  slide  down  if 
the  wall  should  give.  In  any  computation,  regard  0  as  the  intersection  of 
the  plane  of  rupture  with  the  outer  slope  of  the  rip-rap,  •<  ihat  LO  is  the 
horizontal  through  O,  which  thus  determines  h^. 

Assuming  the  load  on  NQ,  W  =  800  Ib.  per  sq.  ft.,  we  find  hr  =  W  /Wi  =  8'. 
Also,  since, 


W 

W" 


W 
W 


wi  hi  =    800  +  500  =  1300  /.  h"    =  20', 
=  1300  +  390  =  1690  /.  h'"  =  26', 


simple  substitution  in  the  formulas,  gives: 

Ei  =  1500  Ibs.,  ai  =  2.30', 

E2  =  4640  Ibs.,  02  =  2.87', 

Ez  =  3470  Ibs.,  a*  =  4.73'. 


52,53]  PASSIVE    THRUST  79 

Whence,  taking  moments  about  A,  the  arms  of  Elf  E2,  E3,  being  18.30',  12.87', 
4.73',  and  dividing  the  sum  of  the  moments  by  the  sum  of  the  thrusts,  E  = 
9610  Ibs.,  it  is  found  that  the  resultant  E  acts  10.8'  above  A  or  very  slightly 
above  the  center  of  AN.  This  result  follows  because  of  the  comparatively 
small  thrust  of  the  rip-rap.  If  earth  was  to  replace  the  rip-rap,  E3  would  be 
three  times  the  value  given.  To  justify  the  computation  of  £2,  as  given,  ob- 
serve, since  OP  is  the  plane  of  rupture  for  the  middle  stratum,  that  the  resulting 
thrust  on  the  vertical  plane  through  0,  for  the  height  of  this  stratum,  is  trans- 
mitted through  the  intervening  earth  and  rip-rap,  to  the  plane  LM  and  is 
thus  the  thrust  Ez  on  LM. 

A  similar  computation  to  the  above  can  now  be  made  for  low  water,  the 
resultant  on  the  base  of  the  wall  ascertained  and  the  result  compared  with 
the  preceding. 

It  is  advisable,  when  practicable,  to  tilt  the  base  of  sea  walls;  otherwise, 
the  probable  small  resistance  to  sliding  on  the  base  may  call  for  a  very  thick 
wall. 

53.  Passive  Thrust.  The  assumptions  in  the  case  of  pas- 
sive thrust  (or  resistance)  are  exactly  those  of  Art.  36,  where 
one  solution  is  given,  the  angles  <p  and  <p'  will  change  sign  and 
another  solution  can  be  effected  by  an  analysis  similar  to  that 
of  Arts.  38-39.  The  final  result  alone  will  be  given  here. 

In  Fig.  27,  where  AB  represents  the  inner  face  of  the  wall, 
BC  the  earth  surface,  draw  AD  to  the  left  of  AB,  making  the 
angle  <p  with  the  horizontal  and  produce  CB  to  meet  AD  at  D. 
Lay  off  BAM  =  <p  -\-  <p't  to  the  right  of  AB  and  from  M,  on 
BC,  draw  a  tangent  MT  to  the  semi-circle  described  upon  DB 
as  diameter.  Lay  off  MC  =  MT,  draw  CI  II  AM  to  inter- 
section 7  with  DA  produced.  Lay  off  on  ID,  IL  =  1C,  then 
if  LC  is  drawn,  the  passive  thrust,  directed  as  indicated  in  Art. 
36,  is  given  by  w  times  the  area  of  the  triangle  CIL  and  AC  is 
the  plane  of  rupture.  This  represents  the  resistance  of  the 
wedge  of  rupture  ABC  to  being  pushed  up  the  planes  AB  and 
AC  by  a  force,  equal  and  directly  opposed  to  this  resistance, 
acting  to  the  right  on  the  wall,  as  shown.  It  is  very  much 
larger  than  the  active  thrust,  particularly  when  AB  is  nearer 
the  vertical  through  A  or  to  the  right  of  it,  when  L  is  often 
found  above  D.  It  is  assumed  that  the  force  on  the  wall  acting 
-to  right  is  applied  at  y£  its  height  above  A,  otherwise  the 
i  earth  resistance  will  not  increase  from  zero  at  B  uniformly 
downwards  and  a  strict  solution  is  not  given  by  the  construction. 


80 


NON-COHESIVE    EARTH.       ANALYTICAL    METHODS 


54.  Ellipse  of  Stress.  It  is  a  well  known  principle,  that 
for  every  condition  of  stress  parallel  to  one  plane,  there  are 
two  planes  at  right  angles  to  each  other  on  which  the  stresses 
are  wholly  normal.*  Such  stresses  are  called  the  principal 
stresses  and  their  directions  the  principal  axes  of  stress. 

In  Fig.  28,  let  AB  and  BC  be  the  planes  at  right  angles  to 
each  other  on  which  the  stresses  are  wholly  normal.  The  figure 


FIG.  28 


FIG.  29 


is  supposed  to  represent  a  cross-section  ABC  of  a  right  prism, 
of  length  unity  at  right  angles  to  the  plane  of  the  paper,  with 
the  unit  stresses,  a,  b  and  r,  acting  on  the  three  faces  AB,  BC 
and  CA,  the  stresses  all  acting  parallel  to  the  plane  of  the  paper. 
The  lengths  AB,  BC,  CA,  will  be  taken  as  infinitesimal,  so  that 
the  unit  stress  on  each  face  can  be  regarded  as  uniformly  dis- 
tributed or  of  uniform  intensity  over  that  face. 
Let  a  =  normal  unit  stress  on  face  AB, 

b  =  normal  unit  stress  on  face  BC, 

r  =  unit  stress  on  face  AC, 

x  and  y,  components  of  r  perpendicular  to  BC  and  A  B 
respectively, 

6  =  angle  BAG. 
Assume  a  numerically  greater  than  b. 

*Rankine's  Applied  Mechanics,  p.  98. 


54] 


ELLIPSE    OF     STRESS 


81 


The  principal  stresses  a  and  b  are  taken  as  causing  com- 
>ression  on  the  faces  AB,  BC.  If  either  one  is  reversed  in 
lirection,  its  sign  will  be  changed  in  the  formulas  below. 

The  three  forces  a. AB,  b.BC,  r.AC,  are  in  equilibrium; 

ice  balancing  components  parallel  to  AB  and  BC  respectively, 

x.  AC  =  b.  BC  =  b.  AC  sin  6 
y.  AC  =  a.  AB  =  a.  AC  cos  0 

.'.    x/b  =  sin  0,  y/a  =  cos  0      .      .     .      .    (29) 


x2       y2 

-  +  :-  =  sin*  0  +  cos2  0  =  i 
b-       a2 


(30) 


This  is  the  equation  of  an  ellipse,  called  the  ellipse  of  stress, 
ig.  29,  in  which  the  components  x,  y  oi  r  are  shown  as  the  ordi- 
ates  to  the  ellipse  at  the  point  R.     We  have,  r  =  V#2  +  y2  = 
R,  acting  upwards  for  the  case  shown,  EO  equal  and  parallel 
,o  a,  DO  equal  and  parallel  to  b  and,  if  we  draw  OM  \\  AC,  then 

OD'  =  e. 

With  center  O,  construct  circles  with  radii,  a  and  b  and  draw 
_L  OM  to  meet  the  outer  circle  at  N.     The  stress  r  =  OR, 
hus  acts  on  a  plane  parallel  to  OM,  whose  normal  is  ON. 

The  obliquity  of  the  stress  is  defined  to  be  the  angle  RON, 
t  the  stress  makes  with  the  normal  to  the  plane  on  which 
t  acts. 

It  is  shown  in  Analytic  Geometry,  that  if  from  the  point 
here  ON  cuts  the  inner  circle  a  line  SR  is  drawn  parallel  to  OE 
d  from  N,  a  line  NI  parallel  to  OD,  their  intersection  R  is 
point  on  the  ellipse,  corresponding  to  the  eccentric  angle  POE 
hich  equals  MOD'  =  0,  as  marked,  the  corresponding  sides 
ing  perpendicular.     Any  radius  vector  to  the  ellipse  gives  the 
Btress  on  a  plane  that  can  be  determined;    thus  if  0  R  is  the 
stress,  draw  R  N  \\  0  D  to  N,  the  intersection  with  the  Outer 
'circle,  then  O  N  is  the  normal  to  the  plane  on  which  0  R  acts. 
Drawing  0  M  _L  0  N,  it  follows  that  the  stress  0  R  acts  on  a 
plane  parallel  to  O  M  and  that  its  obliquity  is  R  0  N. 

A  useful  method  of  drawing  an  ellipse  is  as  follows:  Suppose 
the  axes  o  x,  o  y,  Fig.  29,  drawn  at  right  angles  to  each  other. 
Lay  off  on  the  straight  edge  of  a  sheet  of  paper,  distances  F  R  = 


82  NON-COHESIVE  EARTH.      ANALYTICAL  METHODS 

b,  R  Q  =  a,  then  always  keeping  the  points  F  and  Q  on  the  axes 
oy,  ox,  as  the  line  FQ  is  shifted,  the  point  R  will  describe  an 
ellipse  with  semi-axes  a  =  O  E,  b  =  0  D. 

The  proof  is  easy.    Thus  calling  x,  y,  the  coordinates   of 
R  and  noting  that  Z  /  F  R  =  Z  S  RQ, 

sin  IFR  =  -,  cos  IFR  =cos  S  RQ  =  - 
b  a 

•••$+$-  I 

which  is  the  equation  of  an  ellipse,  in  fact  of  the  ellipse  of  stress. 
Comparing  the  equations  above  the  last  with  eqs.  29,  we 
see  that 

Z  IFR  =  6  =  NOF, 

hence,  P  being  the  intersection  of  the  normal  ON  with  FQ, 
OP  =  PF;  also,  since  POQ  =  90°  -  6  =  PQO  /.  0  P  =PQ 
=  PF. 

The  external  angle  F  P  N  of  the  triangle  P  0  F  is  equal  to 
the  sum  of  the  two  interior  opposite  angles,  P  0  F,  P  F  0; 

:.  FPN  =  26. 

Some  important  conclusions  can  now  be  drawn. 
Since  OP  =  PQ  =  PF,  b  =  F  R,  a  =  RQ, 

b  •+  R  P  =  a  -  R  P  :.  R  P  =  -  (a  -  b); 


OP 

.'.  OP  =  -  (a  +  b),  R  P  =  -  (a  —  &),  are  constant  for  any 

point  R  of  the  ellipse. 

Hence  if  the  plane  0  M,  on  which  the  stress  0  R  can  be 
supposed  to  act,  rotates  about  0,  the  point  P  on  its  normal, 
describes  a  circle  about  O  with  a  radius  }4  (a  +  b)  and  the 
point  R  describes  a  circle  about  P  with  a  radius  }4  (a  —  b), 
with  the  condition  that  O  P  and  P  R  shall  always  be  equally 
inclined  to  0  F  or  that  P  F  shall  be  made  equal  to  O  P. 


4-56] 


ELLIPSE    OF    STRESS 


83 


Thus  if  the  unit  stress  on  a  plane  _L  0  P'  is  desired,  lay  off 
>  P'—  0  P=  -  (a  +  5);  with  P'  as  a  center,  describe  an  arc 
radius  P'O,  cutting  0  y  at  £*  and  on  P'F'  lay  off  P'#'= 
=  -  (a  —  &).  Then  #'  is  a  point  on  the  eUipse  of  stress 
.nd  C  7?'  represents  the  unit  stress  on  the  plane  whose  normal 

3  OP'. 

55.  Normal  and  tangential  components  of  the  unit  stress  on 
'ie  face  A  C,  Fig.  28,  which  is  parallel  to  0  M,  Fig.  29.  On 
ropping  a  perpendicular  R  T  upon  0  N,  Fig.  29,  we  have  the 
omponent  of  OR  normal  to  the  plane  OM  =  O  T  and  the  com- 
onent  parallel  to  OM  =  TR. 

-(a  +  V)  +     (a-b)  cos  26  .  .  (31) 


.  (32) 


TR  =  PR  sin  OPR  =  -(a-b)  sin  20 


Note  that  6  =  angle  made  by  the  plane  A  C,  Fig.  28,  or  its 
arallel  OM,  Fig.  29,  with  the  minor  axis  of  the  ellipse  of  stress. 

In  case  the  principal  stress  b  is  tensile,  its  direction,  as  well  as  that  of  x, 
reversed  in  Fig.  28.  The  unit  normal  stress  b  must  now  be  considered  in- 
•insically  negative  (x  likewise  taking  the  nega- 
ve  sign  from  the  first  equation  of  Art.  54)  and 
ic  results  of  Arts.  54-55,  will  all  be  found  to 
pply.  The  student  is  advised  to  re-read  these 
rticles  in  connection  with  Fig.  30,  to  verify  this 
atement. 

Since    b   is    negative,    RP  =  ^2  (a  —  b),   will 
2  greater  than  OP  =  ^  (a  +  b)  and  the  point 

will  fall  on  the  other  side  of  OF  from  ON. 
he  unit  stress  on  the  plane  OM,  whose  normal  is 
is  as  before,  OR  in  amount  and  direction, 
erify  that  its  normal  and  tangential  compo- 
ants  are  correctly  given  by  (31),  (32)  of  Art.  55. 

In  the  remaining  articles  of  this  chapter,  the  principal  stresses  a  and  b 
ill  always  be  considered  compressive. 

56.  The  preceding  conclusions  hold  for  any  state  of  stress 
i  a  solid  body  and  also  for  a  granular  mass ;  but  with  the  con- 


Q     D     X 


FIG.  30 


84  NON-COHESIVE   EARTH.      ANALYTICAL   METHODS 

dition  that  the  obliquity  RON  (or  the  angle  that  the  stress 
O  R  makes  with  the  normal  to  the  plane  on  which  it  acts)  shall 
not  exceed  <p,  the  angle  of  friction. 

The  angle  R  0  N,  Fig.  29,  is  a  maximum  when  P  R  is  per- 
pendicular to  OR,  in  which  case  a  circle  described  with  P  as 
center  and  PR  radius,  will  be  tangent  to  0  R. 

.'.  sin  (p  —  sin  NOR  =  -    -  =  - 

OP      a  +  b 

For  any  other  position  of  R  (on  the  circle  mentioned)  the  per- 
pendicular dropped  from  P  upon  O  R  will  be  less  than  P  R  and 
sin  NO  R  will  thus  be  less  than  the  value  just  given.  We  derive 
at  once  from  the  last  formula, 

b       i  —  sin  (p 

~  =       ,     .          (33) 

a       i  -f-  sin  <p 

This  gives  the  ratio  of  the  two  principal  stresses  in  the  case 
of  earth  devoid  of  cohesion  and  having  at  the  point  of  the  mass 
considered  the  angle  of  friction  <p.  This  angle  is  generally  as- 
sumed constant  throughout  the  mass,  as  will  be  done  in  what 
follows. 

57.  In  the  next  figure,  31,  the  free  surface  of  earth  is  taken 
as  plane  and  making  the  angle  i  with  the  horizontal.  The 
infinitesimal  wedges  of  rupture  are  represented  by  the  triangles 
A  B  C,  having  one  face  B  C,  vertical  and  one  face  parallel  to  the 
free  surface;  cases  (a)  and  (b)  referring  to  active  thrust  where 
the  wedge  is  about  to  slide  down  A  C,  cases  (c)  and  (d),  to  pas- 
sive thrust  or  resistance,  where  the  wedge  is  on  the  point  of 
sliding  along  the  plane  C  A  in  the  direction  C  A . 

The  reaction  of  the  earth  below  A  C,  has  a  normal  component 
and  one  parallel  to  A  C.  The  latter  or  tangential  component 
in  cases  (a)  and  (b),  is  directed  up  the  plane  A  C,  but  in  cases 
(c)  and  (d),  its  direction  is  from  A  to  C,  hence  r" ,  the  unit  re- 
action, is  drawn  as  shown  for  the  four  cases,  always  making 
the  angle  (p  with  the  normal  to  the  plane  of  rupture  A  C* 

*  It  may  be  remarked  that  case  (d)  always  corresponds  to  incipient 
motion  up  the  plane  CA;  but  in  case  (c),  a  wedge  of  rupture  can  be  formed 
when  the  point  A  lies  below  the  level  of  C,  the  extreme  limiting  position  cor- 


ELLIPSE    OF    STRESS 


85 


The  vertical  unit  stress  on  AB  (=  weight  of  vertical  prism 
tanding  on  AB  when  AB  —  i)  =  r  =  wx  cos  i,  where  x  = 
ertical  depth  of  AB  below  the  free  surface  and  w  =  weight 
er  cubic  unit  of  the  earth.  The  unit  pressure  on  the  vertical 
lane  B  C  is  r'  and  in  an  unlimited  mass  of  earth,  subjected  to 
o  external  force  (as  the  friction  along  a  wall,  etc.)  but  its  own 
eight,  r'  acts  parallel  to  AB  and  to  the  free  surface,  Art.  n. 

Since  either  r  or  r'  acts  parallel  to  the  plane  on  which  the 
ther  acts,  they  are  conjugate  unit  stresses,  with  a  common 


FIG.  31 


,  since  each  makes  the  angle  i  with  the  normal  to  its 
plane,  Art.  n. 

In  cases  (a)  and  (b)   of  active  thrust,  the  weight  of  earth 

h)ver  A  B,  causes  the  tendency  to  slide  down  A  C,  giving  an 

iarth  thrust  opposed  in  direction  to  r'  as  drawrf^  If  the  faces 

£  C  are  placed  together,  the  unit  thrust  of  magnitude  r'  in  case 

|ta),  must  exactly  equal  and  be  opposed  to  the  thrust  of  case 

\b)  or  equilibrium  will  be  impossible. 

In  cases  (c)  and  (d),  the  unit  thrust  /,  refers  to  some  external 
iorce,  which  is  sufficiently  great  to  cause  incipient  sliding  in 

responding  to  C  A  parallel  to  AB.  Hence  in  case  (c),  according  as  A  is  above 
or  below  the  level  of  C,  the  incipient  motion  is  up  or  down  the  plane  CA. 
The  student  is  cautioned  that,  for  any  one  of  the  cases,  the  unit  stresses 
r,  /,  r",  acting  on  a  wedge  of  rupture,  are  not  in  equilibrium,  but  that  r.  AB, 
r'.  B  C,  r".  CA,  do  constitute  a  system  of  concurrent  forces  in  equilibrium. 
Also,  it  may  be  noted,  on  assuming  B  C  the  same  in  cases  (c)  and  (d),  that 
:when  A  is  below  the  .level  of  C  in  case  (c),  r.  AB  is  much  larger  for  this  case 
ithan  for  case  (d),  which  explains  why  r'  can  be  the  same  for  both  cases,  for  the 
same  value  of  x. 


86       NON-COHESIVE  EARTH.   ANALYTICAL  METHODS 

the  direction  C  A .  This  means  that  the  wedge  ABC  does,  not 
slide,  but  that  the  slightest  increase  in  /  will  cause  actual  sliding. 
If  we  suppose  the  faces  B  C  in  cases  (c)  and  (d)  to  coincide, 
the  magnitudes  of  /  for  the  two  cases  must  be  equal,  since  the 
thrust  for  one  wedge  must  equal  the  reaction  of  the  other. 

In  all  four  cases,  the  three  forces  r .  A  B,  rf .  B  C  and  r" .  A  C, 
must  be  in  equilibrium  and  the  analytical  solution  is  dependent 
upon  this  fact.  In  the  ellipse  of  stress,  however,  it  must  be 
borne  in  mind  that  a  radius  vector  as  0  R,  gives,  not  the  total 
stress  on  a  plane  parallel  to  0  Mj  but  the  unit  stress  on  this 
plane. 

58.  Conjugate  Stresses.  In  Fig.  29,  suppose  r  =  0  R  to 
represent  a  unit  stress  acting  on  a  plane  0  M,  then  its  conjugate 
stress  /  must  act  in  the  direction  0  M.  Take  r'  =  0  R'  to 
representative  magnitude  of  this  stress  which  acts  on  a  plane 
OR.  Since  the  stresses  are  conjugate,  they  both  have  the  same 
obliquity  *  =  N  0  R  =  P'O  R'. 

By  Art.  54,  we  have  0  P  =  OP',  PR  =  P'R'.  To  find  the 
ratio  of  the  magnitudes  of  r'  and  r: 

(1)  Let  us  suppose  both  0  P  and  0  P'  of  Fig.  29  to  be  laid 
off  along  the  same  straight  line  OJ  from  0  to  P,  Fig.  3  2  (a)  and 
that  lines  0  R",  0  R,  are  drawn  making  the  angles  <p  and  i  re- 
spectively with  0  P  J.     On  dropping  a  perpendicular  P  R"  upon 
0  R"  from  P,  use  P  as  a  center,  P  R"  as  radius,  to  describe  the 
semi-circle,  which  meets  0  R  produced  at  Rf  and  0  P  J  at  /  and 
/.     Then  P  R"  is  equal  in  length  to  P  R  of  Fig.  29,  since  for 
the  maximum  obliquity  <f>,  P  R"  is  perpendicular  to  0  R",  Art. 
56  and  P  R  for  any  point  on  the  ellipse  is  of  constant  length. 
It  follows  that  the  triangles  0  P  R  of  Figs.  ?p  and  3 2 (a),  are 
equal;  also  the  triangles  0  P' Rf  of  both  figures  are  equal,  so 
that  the  magnitudes  of  0  R,  0  R'  are  the  same  in  either  figure, 
though  the  unit  stresses  represented  by  them  in  Fig.  32  (a),  are 
not  in  true  directions.     This  construction  has  been  made  prin- 
cipally as  an  aid  to  the  better  understanding  of  the  next  two 
constructions. 

(2)  Let  us  suppose  that  any  convenient  length  for  0  P  in 
Fig.  3  2  (a)  is  assumed  and  the  remaining  construction  effected 


58] 


CONJUGATE    STRESSES 


87 


as  before.  Evidently  the  figure  will  be  similar  to  the  former 
one,  the  ratio  0  R'/O  R  will  be  the  same  for  either  figure,  so 
that  the  ratio  of  the  conjugate  stresses  can  be  found  from  this 
construction  without  knowing  the  magnitude  of  either  one 
of  them.* 

(3)  Suppose  the  common  obliquity  and  the  magnitude  of 
one  of  the  conjugate  stresses,  say  r,  to  be  known.    Draw  the 


R" 


o 

I                    P,P' 
(a) 

J          O 

FIG.  32 

i                P,P'              J 

w 

lines  O  R',  0  R",  in  Fig.  3 2 (a),  making  the  angles  i  and  v>  with 
0  /  and  lay  off  to  scale  on  0  Rf,  0  R  =  r.  Since  it  happens  in 
Fig.  29  that  0 Rf  >  OR,  find  by  trial  a  point  P  on  0 /,  Fig. 
3 2 (a),  such  that  the  circle  with  P  as  Center,  PR  radius,  will 
be  tangent  to  0  R",  that  circle  being  taken  which  has  the  greatest 
radius.  The  point  P  thus  found  evidently  coincides  with  the 
P  of  the  first  construction  and  the  figure  coincides  with  the  pre- 
ceding throughout.  The  magnitude  of  r'  is  given  by  the  length 
OR'  to  scale. 

As  previously  found, 

OP  =  i  (a  +  ft),  PR  =  P*R'  =  I  (a  -  ft)  =  PI  =  PJ. 

2  2 

.'.  OJ  =  a,  01  =  b 

The  ratio  //r  of  the  conjugate  stresses  of  Fig.  31,  is  readily 
computed.  Thus  if  T,  Fig.  3 2 (a),  is  the  foot  of  the  perpendicular 
from  P  on  OR, 

OR'      OT  +  R'T      OT  + 


OR       OT-RT       OT- 

O  P  [cos  i  +  ^sin*  <?  -  sin9  i] 
O  P  [cos  i  —   ^sin2  <p  —  sin*  i] 


*  Rankine  gives  this  construction  in  his  "  Civil  Engineering,"  p.  320. 


88  NON-COHESIVE   EARTH.      ANALYTICAL  METHODS 

/ 

r'      cos  i  +  "V  cos*  i  —  cos2  <p 


(34) 


f       cos  i  —  \  cos2  i  —  cos2  <p 
This  gives  the  ratio  r' '/r  for  cases   (c)  and  (d),  Fig.  31,  corre- 
sponding to  a  passive  thrust,  /  >  r. 

For  active  thrust,  cases  (a)  a.nd  (ft),  Fig.  31,  /<  ;-.  Fig.  29 
will  apply  to  this  case  if  accents  are  removed  from  P',  R',  and 
applied  to  P,  R,  so  that  now  0  R'  <  0  R.  Fig.  32(6)  represents 
the  corresponding  abbreviated  construction,  0  R  being  laid  off, 
to  some  scale  =  r ,  and  a  point  P  on  0  J  found  by  trial,  such 
that  a  circle  with  P  as  center,  P  R  radius,  will  be  tangent  to 
OR",  that  circle  being  taken  which  has  the  least  radius. 

As  before,  r  =  0  R  —  vertical  unit  pressure  on  a  plane 
parallel  to  the  free  surface,  at  the  depth  considered,  whence 
O  R'  to  the  same  scale  will  represent  r' '. 

The  ratio  for  active  thrusts  is  thus,  Fig.  32(6), 
/       OR'       cos  i  —  V  cos2  i  —  cos2  <p 

r       OR       cos  i  +  V  cos2  i  —  cos2  <p 

At  the  vertical  depth  x,  r  =  wx  cos  i;  on  substituting  this 
value  in  the  last  two  equations  we  are  conducted  to  Rankine's 
formulas  for  r1 ',  the  unit  pressure  on  a  vertical  plane,  at  a  vertical 
depth  x,  the  direction  of  the  pressure  being  parallel  to  the  free 
surface. 

The  ratio  r'/r,  for  either  active  or  passive  thrust,  can  either 
be  computed  from  the  above  formulas  or  the  values  OR,  OR' 
can  be  scaled  from  the  figures  and  the  ratios  OR' /OR,  computed. 

Note,  that  from  one  of  the  similar  figures  (a)  and  (£),  Fig. 
32,  the  other  can  be  obtained  by  simply  interchanging  R'  and 
R,  so  that  hereafter  only  one  figure  need  be  dr$wn  to  find  the 
ratio  r' jr. 

After  the  circle  of  Fig.  32  has  been  determined  as  above, 
we  may  regard  OR  R'  as  no  longer  making  a  constant  angle  i 
with  0  J,  but  as  having  any  position  from  0  J  to  OR".  Then, 
for  any  one  of  these  positions  of  ORR',OR  and  ORf  will  represent 
conjugate  stresses  with  a  common  obliquity  ROJ.  Thus,  if 
A  ROP  in  Fig.  29  corresponds  to  and  is  thus  equal  to  A  R  0  P 
of  Fig.  3 2 (a),  then  as  shown  above,  A  R'OP'  in  Fig.  29  is  equal 


58-60]  CONJUGATE    STRESSES  89 

to  A  R'OP'  in  Fig.  3 2 (a);  whence  stress  OR,  Fig.  29,  acting 
on  plane  OR'  is  conjugate  to  stress  OR'  acting  on  plane  OR 
and  both  stresses  have  the  common  obliquity  R  OP  =  R'OP' . 

When  ORR',  Fig.  3 2 (a),  coincides  with  OJ,  the  conjugate 
stresses  OR,  OR',  reduce  to  the  principal  stresses  01,  OJ.  When 
OR  R'  is  revolved  to  the  position  OR" ,  then  the  conjugate 
stresses  are  both  equal  to  OR".  Since  the  common  obliquity  is 
now  the  maximum  <p,  the  unit  stress  OR"  is  that  on  either  plane 
of  rupture. 

59.  For  active  thrust,  on  putting, 

1  cos  i  —  \  cos2  i  —  cos2  <p 
K  =  -  cos  i .  , 

2  cos  i  -h  \  cos2  i  —  cos2  <p 

we  have  from  (35),  after  substituting,  r  =  wx  cos  i, 

rr  =  2  K  w  x (36) 

which  is  identical  with  (26),  Art.  49,  when  x  =  h.  It  gives  the 
unit  stress  or  intensity  of  the  conjugate  stress  r'  on  a  vertical 
plane,  at  the  depth  x.  Hence  the  total  earth  thrust,  acting 
parallel  to  the  surface,  on  a  vertical  plane  of  depth  h,  is, 

•*-'  J  o  '  '  '        \O  I  / 

which  is  equivalent  to  (25),  Art.  48,  proved  in  an  entirely  different 
manner. 

The  corresponding  formulas  for  passive  thrust  or  resistance 
are  of  the  same  form  as  (36)  and  (37),  but  the  value  of  K  is 
different.  Let  the  student  write  the  value  of  K. 

From  (34)  and  (35),  substituting  r  —  wx  cos  <p,  we  have, 
when  i  =  <p,for  both  active  and  passive  thrusts,  r'  =  wx  cos  <p=r. 


When  i  =  o,  the  active  unit  thrust,  r'  =  w  x 


the  passive  unit  thrust,  r'  —  w  x 


i  +  sin  <p* 
i  -\-  sin  (p 


i  —  sin  <p 

60.  An  application  will  now  be  made  of  the  ellipse  of  stress 
to  finding  the  planes  of  rupture  and  the  amount  and  direction 
of  the  active  pressure  on  any  plane,  at  a  given  depth  below 
i  the  free  surface  of  earth,  which  will  be  supposed  to  make  an 
angle  i  with  the  horizontal. 


90 


NON-COHESIVE  EARTH.      ANALYTICAL  METHODS 


In  Fig.  33,  draw  OG  perpendicular  to  the  free  surface,  then 
the  vertical  pressure  per  square  unit  on  a  plane  OA,  parallel 
to  the  surface,  at  a  vertical  depth  x,  =  wxcosi  =  w.OG  = 
w  .  OR,  where  OR  is  vertical  and  equal  to  O  G.  Hence  reserving 
the  factor  w,  the  point  R  will  be  one  point  on  the  ellipse  of 
stress  corresponding  to  w  =  i.  Now  the  pressure  on  a  plane  of 


FIG.  33 

rupture  through  0  makes  an  angle  <p  with  its  normal.  Suppose 
this  plane  revolved  around  0  until  its  normal  takes  the  position 
OG  and  lay  off  the  angle  GOR0  =  <p,  to  the  right  of  OG.  Then 
P  and  Rj  having  the  same  meaning  as  in  Fig.  29,  remembering 
that  OP  and  PR  are  constants,  find  by  trial,  the  point  P  on  0  G, 
such  that  an  arc  of  a  circle  with  P  as  center  and  PR  as  radius, 
will  be  tangent  to  OR0.  Drawing  PR0  _L  OR,,  the  obliquity 
POR0  is  a  maximum  (Art.  56)  and  equal  to  <p,  the  angle  of 
friction. 

Lay  off  PGH  =  PR  =  PR0,  also  extend  PR  in  either 
direction.  Now  since  R  is  one  point  on  the  ellipse  of  stress 
and  since  by  Art.  54,  OP  and  PR  must  make  equal  angles  with 
its  major  axis,  lay  off  PR F  =  OP,  whence  £POF=  LPFO, 
and  draw  OF  to  give  the  direction  of  the  major  axis.  On  drawing 
OD  Q  _L  0  F,  by  the  method  of  constructing  an  ellipse,  Art.  54, 


>0]  CONJUGATE    STRESSES  91 

ve    have    RF  =  b,    RQ  =  OH  =  a.     Otherwise,   as    hitherto 
Droved, 

a  +  b      a  —  b 


a  +  b      a  —  b 
RF  =  PF  -  PR  =  -  -  =  b. 

2  2 

Next  lay  off  on  OF,  OE  =  a  =  OH  and  on  OQ,  OD  = 
=  b.  Also  mark  on  the  straight  edge  of  a  sheet  of  paper 
or  other  straight  edge)  the  points  Q,  R,  F\  then  on  shifting  the 
itraight  edge  so  that  the  outer  points  always  remain  on  the 
ixes  OQj  0  F,  produced,  the  inner  point  will  describe  the  ellipse 
)f  stress. 

PLANES  OF  RUPTURE.  One  plane  of  rupture  can  be  found  as  follows: 
With  center  0  and  radius  ORo  describe  an  arc  of  a  circle,  cutting  the  ellipse 
it  K.  Draw  P^  _L  OK  and  lay  off  PoK  =  PR<,  .'.  A  P0OK  =  A  PORo 
md  Z  P0OK  =  Z  PORo  =  <p.  Thus  it  appears  that  A  PORo  is  the  revolved 
position  of  A  P0OK,  which  is  in  true  position,  OP0  being  the  normal  to  the 
blane  of  rupture  on  which  the  unit  stress  is  iv .  0  K,  this  stress  making  the 
ingle  P0OK  =  <p  with  the  normal  to  the  plane. 

An  alternative  construction  is  available.  Draw  a  perpendicular  through 
'K  to  the  major  axis  OE  to  meet  a  circle  described  with  0  as  center  and  OE  =  a 
is  radius,  at  M  and  Mi,  then  M O  coincides  in  position  with  P00,  Art.  54. 

If  M  K  Mi  meets  the  ellipse  to  the  right  of  OE  at  KI,  then  from  symmetry, 
'Z  KiOMi  =  KOM  =  <p\  whence  OMi  is  the  normal  to  the  second  plane  of 
rupture  on  which  the  unit  stress  is  w  .  OK\  =  w  .  OK. 

From  the  figure,  it  may  be  stated  as  a  proposition,  that  the  planes  of 
rupture  make  equal  angles  with  and  on  either  side  of,  the  major  axis,  for  the 
of  active  thrust  here  considered. 

To  avoid  confusing  the  figure,  the  planes  of  rupture  are  not  drawn.  One  is 
perpendicular  to  OM,  the  other  to  OM\  nearly  coinciding  with  OK\  and  OK, 
respectively  for  the  i  and  <f>  assumed.  This  construction  for  planes  of  rupture 
is  not  so  accurate  as  that  given  hitherto  or  the  one  to  follow,  Art.  65,  since 
the  ellipse  is  not  generally  so  accurately  drawn  that  the  point  K  is  fixed  with 
certainty. 

To  find  the  unit  stress  at  0  on  a  plane  parallel  to  AB,  draw 
OP'  _L  AB  and  make  OP'  =  OP.  Let  i  be  thejoot  of  a  per- 
pendicular from  P'  on  OE.  Lay  off  along  OE,  12  =  Oi,  draw 
Pf2  and  on  it  lay  off  P'R'  =  PR.  It  follows  from  Art.  54 
that  w .  R'O  is  the  unit  stress  at  O  on  a  plane  parallel  to  AB, 

• 


92       NON-COHESIVE  EARTH.   ANALYTICAL  METHODS 

R'O  being  the  direction  of  the  stress;  since  for  any  point  R' 
on  the  ellipse  we  must  have,  OP'  =  OP,  P'R'  =  PR,  QPf  and 
P'R'  making  equal  angles  with  the  major  axis.* 

The  total  thrust  on  the  plane  AB  in  an  unlimited  mass  of 
earth  is  found  as  follows :  The  thrust  at  A  =  w .  R'O  and  it 
acts  ||  R'O.  Since  the  thrust  varies  uniformly  with  the  depth 
from  the  surface,  if  we  lay  off  AT  ±.  AB,  of  length  AT  =  OR' 
and  draw  BT,  w  X  area  triangle  ABT  will  give  the  total  thrust 
on  the  plane  AB.  Its  resultant  acts  parallel  to  R'O  at  C,  where 
AC  =  yzAB. 

When  AB  is  the  inner  face  of  a  retaining  wall  and  AB  makes 
a  greater  angle  with  the  vertical  than  the  plane  of  rupture 
_L  OMi  (OK  very  nearly),  this  construction  applies  to  finding 
the  thrust  on  the  wall,  provided  </  _^_  <p.  As  drawn,  AB  is  less 
inclined  than  the  normal  to  OM\  to  the  vertical;  hence  the 
construction  of  Arts.  21  or  39  must  be  used.  As  walls  are 
generally  rough  or  stepped  on  the  inner  face,  it  rarely  happens 
that  <p'  <  <p.  It  sometimes  occurs  by  design  in  the  case  of 
experimental  walls.  In  this  case  if  the  construction  of  Fig.  33 
gives  R'O,  making  a  greater  angle  than  <p  with  the  normal  to 
the  wall  AB,  then  the  construction  of  Art.  21  or  Art.  39  applies. 

The  stress  on  the  plane  AB,  Fig.  33,  can  be  found  also  by  a 
method,  which  avoids  drawing  the  ellipse  of  stress.  Thus,  let 
a  vertical  plane  through  A  intersect  the  free  surface  at  L  and 
denote  AL  by  x.  Then  the  unit  stress  r'  at  A,  acting  parallel 
to  the  surface,  is  given  by  the  construction  Fig.  32(6).  Thus 
when  OR  =  wx  cos  i,  the  construction  gives,  /  =  ORf.  The 
total  stress  on  the  vertical  plane  AL,  Fig.  33,  is  thus,  y£  r'x, 
acting  y$  x  above  A  and  parallel  to  the  surface.  Lay  this  off 
to  scale,  along  the  surface,  say  equal  to  LU  ana  irom  U  draw  a 
vertical  line  UV  of  the  length,  to  the  same  scale,  equal  to  the 
weight  of  earth  ABL.  The  resultant  of  these  two  forces,  LV, 
gives  the  amount  and  the  direction  of  the  stress  on  AB.  It 
acts  at  the  point  C. 

61.  When  the  earth  surface  makes  the  angle  $  with  the  horizon- 
tal, the  construction  is  simplified.  OR  being  vertical  and  equal 

*  Compare  with  Howe's  Retaining  Walls,  p.  6. 


o  OG,  the  angle  GOR,  Fig.  34,  is  now  equal  to  <p  and  the  line 
R0  coincides  with  OR.  The  point  P  is  found  at  the  inter- 
action of  a  perpendicular  at  R  to  OR  with  OG  produced,  for 
hen  the  angle  P  OR  is  the  maximum  <p.  As  before,  we  lay  off 
PR  F  =  P  0  to  locate  a  point  F  on  the  major  axis.  Also,  b  = 
R  F,  a  =  OP  +  PR ;  whence  the  ellipse  of  stress  can  be  drawn 
f  desired. 

Since  R  0  makes  the  angle  <p  with  the  normal  to  AO,  AO  is 
ne  plane  of  rupture.     Also,  since  the  conjugate  stress  to  RO 


1] 


STRESS    ON    ANY    PLANE 


93 


FIG.  34 

acts  on  a  vertical  plane  in  the  direction  AO  (see  Fig.  31),  making 
angle  <p  with  the  normal  to  the  vertical  plane,  the  latter  is 
he  second  plane  of  rupture,  or  "  limiting  plane"  of  Art.  27. 
Hence,  when  <p'  2l  <p,  to  find  the  unit  stress  at  A  on  the  inner 
ace  AB   of  a  battered  retaining  wall,   draw  OP'  =  OP  and 
_L  AB.    With  P'  as  a  center  and  P'O  radius,  describe  an  arc 
cutting  OF  produced  at  2.     On  Pf2  lay  off  P'R'  =  PR,  whence 
.  R'O  will  give  the  unit  stress  required  at  A.    Next  lay  off 
AT  JL  AB,  AT  =  OR']    then  w.  area   triangle  ABT  will  give 
the  total  stress  on  AB.     Its  resultant  acts  at  C  (AC  =  y£  AB), 
a  direction  parallel  to  R'O.    As  an  exercise,  take  the  vertical 
height  of  AB  =  i  foot,  to  a  scale  of  5  or  10  inches  to  i  foot  and 
i  assume  w;  =  i;  then  the  thrust  on  AB  =K  AB  .OR'  =  E  =  K. 
Find  the  value  of  K  in  the  following  table,  corresponding  to 


94 


NON-COHESIVE  EARTH.   ANALYTICAL  METHODS 


<P  =  33°  41'  and  a  batter  of  AB  of  2  inches  to  the  foot;  also 
the  value  of  X  =  angle  made  by  E  with  the  normal  to  AB. 

In  this  table, 
a  =  angle  made  by  AB  with  the  vertical, 

Ki  =  K  cos  X  =  component  of  K  normal  to  AB. 

The  values  of  K\  given,  will  be  used  in  the  next  chapter  on  the 
designing  of  walls.  By  the  graphical  method,  the  values  of  K 
can  only  be  counted  on  to  three  significant  figures,  with  a  possible 
error  of  i  or  2  in  the  third  place  and  the  error  in  X  may  reach  15'. 


Batter  of  AB 

a 

A 

K 

Ki 

o' 

I1 

to      
to       

0 

\ 

°oo' 

46 

33 

T.T. 

3  41' 
II 

0.4161 
.4922 

o  .  3462 

.4119 

->' 

to 

28 

^1 

S8 

5716 

4.84Q 

y 

to 

14. 

O2 

^o 

1C 

6SOS 

.  5697 

?' 

to 

18 

26 

-?8 

71:4.7 

.6646 

5' 

to  I   ... 

?? 

VJ 

?6 

on 

.8566 

.7696 

The  values  of  X  and  K  were  computed  from  the  following 
formulas,*  where  e  is  the  angle  made  by  E  with  the  vertical. 


-  (45°-  -)"]=  tow (45°+  -  -  « 


--}  •  -(38) 


eis  found  from  this  formula,  then  X  =  90°  —  (c  +  a). 

I  cos  (<p  —  a)  tan  a 


(39) 


2  COS  (<p  +   € )  COS   a 

Remark.  In  Art.  58,  the  relation  between  the  vertical  unit 
pressure  and  its  conjugate  unit  pressure  parallel  to  the  steepest 
declivity,  was  found.  There  is  a  third  unit  pressure  perpendic- 
ular to  the  plane  of  the  first  two  and  therefore  horizontal. 
Since  it  acts  normally  to  a  vertical  plane,  it  is  a  "principal " 
pressure.  In  the  general  case,  Fig.  33,  this  pressure  at  the 
point  0  acts  at  right  angles  to  the  plane  of  the  paper  and  is 

*  The  formulas  were  derived,  by  first  finding  the  thrust  on  the  vertical 
plane  through  A  and  then  combining  it  with  the  weight  of  earth  between 
that  plane  and  A  B.  The  work  is  too  long  to  give  here.  The  value  of  A  may 
be  directly  derived  from  (40)  of  the  next  article  on  changing  a  to  (—a)  in  that 
equation,  but  the  above  method  is  shorter. 


1,62]  DIRECTION    OF    THE    PRESSURE    ON   A   WALL  95 

bus  conjugate  to  the  greatest  principal  stress  w  X  OE\  hence 
ti  magnitude  it  equals  w  X  OD.  The  ratio  of  the  latter  to 
he  former  is  given  by  eq.  (33)  of  Art.  56  and  the  value  of  this 
hird  principal  stress,  in  Ib.  per  sq.  ft.,  is  also  given  by  OI  to 
cale,  in  Fig.  32(6),  when  OR  =  wx  cos  i,  x  being  the  vertical 
iepth  from  the  surface  to  point  0.  The  state  of  stress  at  0 
n  any  direction  can  be  represented  by  an  ellipsoid,  having  the 
hree  principal  stresses  as  semi-axes. 

62.  Direction  of  the  Pressure  on  a  Wall  Leaning  Towards 
he  Earth.  Leaning  retaining  walls  are  rarely  built  with  the 
nner  face  making  an  angle  over  10°  with  the  vertical.  For 
nclinations  less  than  10°  say,  the  old  assumption,  that  the 
hrust  makes  the  angle  <p  with  the  normal  to  the  wall  when 
'»'  >_  <f>,  or  the  angle  <?'  when  <p  <  <pt  can  be  used  in  getting  an 
,pproximate  value  of  the  thrust  by  previous  methods. 

These  assumptions  must,  of  course,  hold  when  the  wall  is  ver- 
ical,  but  when  the  wall  rests  on  a  bank  sloping  at  the  angle  of 
epose,  the  reaction  of  the  earth  must  be  vertical  and  directed 
jipwards  and  there  can  be  no  horizontal  thrust.  Between  these 
xtremes,  the  direction  of  the  thrust  must  pass  through  all 
ntermediate  positions  as  the  wall  rotates  from  one  extreme  to 
he  other. 

No  simple  solution  is  available,  but  Resal,*  by  a  very  com- 
>licated  analysis,  reaches  the  conclusion,  that  along  the  under 
ide  of  a  wall  leaning  towards  the  earth,  the  unit  pressure  /  on 
,  vertical  plane  Fig.  31,  makes  the  angle  <p  with  the  horizontal; 
icnce  the  conjugate  vertical  unit  pressure  r  acts  on  a  plane 
aclined  at  the  same  angle  <p  with  the  horizontal,  whatever  the 
urface  inclination  may  be.  See  Art.  33,  where  Resal's  funda- 
nental  hypothesis  is  stated,  from  which  this  conclusion  follows. 

Resal's  conclusion  will  be  adopted,  since  it  gives  correct 
esults  for  the  vertical  wall  and  the  one  leaning  at  the  angle  of 
epose,  and  reasonable  results  for  intermediate  positions. 

Fig.  35  shows  a  ready  means  for  finding  the  angle  X,  made 
>y  the  thrust  on  a  leaning  wall  whose  inner  face  is  AB',  with 


Poussee  Des  Terres/'  Vol.  i,  1903. 


96 


NON-COHESIVE  EARTH.   ANALYTICAL  METHODS 


the  normal  to  A  B' .  The  vertical  earth  pressure  at  A  is  unknown, 
since  no  vertical  prism  of  earth  extends  from  A  to  the  surface, 
as  in  the  case  of  the  battered  wall.  Hence  in  Fig.  35  lay  off 
any  vertical  length  OR  to  represent  r  and  proceed  as  in  Fig.  34 
to  lay  off  FOR  =  <p,  draw  PR  J_  OR  and  lay  off  PRF  =  PO. 
The  construction  now  proceeds  exactly  as  before. 

Thus  to  find  the  direction  of  the  thrust  on  AB',  draw  OP' 
±  AB'  and  make  OP'  =  OP;   then  lay  off  P'i  =  P'o,  P'R'  = 


FIG.  35 

PR',   whence  R'O  is  the  direction  of  the  thrust  on  AB'  and  it 
makes  the  angle  X  =  P'OR'  with  its  normal. 

Similarly  R"0  is  the  direction  of  the  thrust  on  AB"  and 
R'"O  (vertical)  the  direction  of  the  thrust  on  AO  as  previously 
seen.  Note  that  R"O  is  the  direction  of  the  active  thrust  on 
AB" ,  corresponding  to  the  wedge  of  thrust  Fig.  31  (a)  and  (b), 
sliding  down  the  plane  AC. 

A  formula  for  tan  \  can  be  easily  deduced  from  the  figure. 

Let  a  =  angle  made  with  the  vertical  AB  by  the  inner  face 

of  the  wall. 

X  =  angle  made  by  thrust  on  wall  with  the  normal  to 
the  wall. 

If  the  component  of  R'O  or  R"0}  etc.,  parallel  to  the  wall, 
acts  down,  X  will  be  regarded  as  +;  when  it  acts  up,  X  will  be 
taken  as  — . 

To  fix  the  ideas,  take  AB'  as  the  inner  face  of  the  wall. 


62,63] 


MOHR'S    CIRCULAR    DIAGRAM   OF    STRESS 


97 


On  drawing  OX  horizontal,  we  have,  XOP'  =  BAB'  =  a. 
Also,  since  PO  =  PF  (by  construction),  .'.  Z  POF  =  Z  PFO 
=  Z  FOX  and  OF  bisects  the  angle  POX  =  90°  +  <?  .*.  FOX 


45°  + 


P'OF  =  (45 


-  +  a)  =  P'20. 
2 


Whence, 


'  =  90°  —  (<p  +  20)  and  on  dropping  the  perpendicular 
R'C  from  R'  upon  OP'  and  remembering  that  OP'  =  OP, 
\P'R!  =  PR,  we  find, 
P'R'  =  PR=OP  sin  <? 

P'C  =  P'R'  cos  [90°  -  (2«  +  *>)]  =  OP  sin  9  sin  (2*  +  ?) 
R'C'  =  P'£'  sm  [90°  -  (2«  +  *>)]  =  OP  «»  *>  cw  (2«  +  ?) 


X  =  tan  R'OP'  = 


OCf       i  —  sin  <p  sin  (2 a  +  <p) 
is  found  by  Resal  in  a  different  way. 


.(4o) 


Ex.     Let  the  student  prove,  by  considering  the  triangle  OP"R"t  that 
;he  formula  applies  when  X  is  negative,  noting  that  tan  (  —  X)  =  —  tan  X. 

The  following  values  have  been  computed  by  the  writer  for 
various  values  of  p. 


30° 


33°  41' 


=40° 


10° 

I5° 
30° 

33°  4i' 

4°o 

50° 

56°  19' 
60° 


+  30° 
+  29°  26' 
+  27°  31' 
+o23°48' 


-  17°  53' 

-  27°  31' 


+  33V 
+  33' 


02' 


+  30°  42' 
+  26°  03' 

-  4°  34' 

-  13°  18' 

-  24°  21' 

-33°  4i'' 


+  40° 
+  39°  09' 
+  35°  56 
+  29°  03' 
-  16°  55' 

-35°  57' 
-40° 


These  values  of  X  will  replace  y  in  previous  graphical  or 
malytical  solutions. 

63.  Mohr's  Circular  Diagram  of  Stress.  We  shall  now 
recur  to  Fig.  32  of  Art.  58.  This  diagram  is  repeated  in  Fig.  36 
ivith  additional  lines.  For  active  thrust,  it  is  well  to  recall  that 
:he  ratio  of  the  conjugate  stresses,  r'/r  =  OR' /OR  =  ORi/OR, 


98 


NON-COHESIVE  EARTH.   ANALYTICAL  METHODS 


if  Z  IORi  =  Z  IOR'.  If  x  is  the  vertical  depth  to  the  point 
considered  and  h  is  the  length  of  the  perpendicular  from  this 
point  to  the  free  surface,  then  r  =  wx  cos  i  =  w  h  and  in  what 
follows,  OR  to  some  scale,  will  be  supposed  to  represent  r  =  w  h. 
Call  p  and  q  the  normal  and  tangential  components  of  r, 
with  respect  to  the  plane,  parallel  to  the  free  surface,  on  which 
it  acts,  Fig.  36,  upper  figure;  also  let  p'  and  q'  represent  the  nor- 
mal and  tangential  components  of  r'  with  respect  to  the  vertical 


R" 


FIG.  36 


plane  on  which  it  acts  .".  p  =  r  cos  i,  q  =  r  sin  i,  p'  =  r'  cos  i, 
qf  =  r'  sin  i.  In  the  lower  figure,  r  and  r'  aie  not  shown  in 
true  directions,  but  since  ROJ  =  R\OJ  =  i,  we  have  for  the 
magnitudes  of  the  components: 

p  =  0N,q  =  NR,  p'  =  ON',  qf  =  N'Ri. 
It  has  been  previously  shown  that, 

01  =  b,  OJ  =  a  .'.  OP  =  ^  (a  +  ft),  P#  =  ^  (a  -  b). 
Let  Z  £//  =  8  :.   Z  £PJ  =  20, 


•: 


3]  MOHR'S    CIRCULAR    DIAGRAM    OF    STRESS  99 

.'.  p  =  OP  +  PN  =  |  (a  +  b)  +  ±  (a  -  b)  cos  26 
q  =  NR  =  -  (a  —  b)  sin  20. 

On  comparing  these  two  expressions  with  (31)  and  (32),  Art. 
5,  and  noting  the  identity,  we  see  that  B  is  the  angle  that  the 
>lane  on  which  r  acts  makes  with  the  minor  axis  of  the  ellipse 
•f  stress.  Similarly,  letting  Z  RJJ  =  6'. 

p'  =  OP  -  N'P  =  ~2  (a  +  b)  +  ~2  (a  -  b)  cos  26' 
q'  =  R^'  =  i  (a  -  b)  sin  26'. 

Thus  0'  corresponds  in  meaning  to  0  of  eqs.  (31),  (32),  Art.  55 
nd  Fig.  29,  and  hence  it  is  the  angle  that  the  plane  on  which  r' 
.cts  makes  with  the  minor  axis.  Since  this  is  a  vertical  plane, 
t  we  rotate  the  lower  figure  about  7,  until  RJ  is  vertical  (or 
?i/  horizontal),  then  the  planes  Ril,  IR,  on  which  r'  and  r 
ct,  will  have  their  true  directions;  vertical  and  parallel  to  the 
tee  surface.*  The  minor  axis  is  parallel  to  01  in  the  revolved 
•osition  and  the  major  axis  is  perpendicular  to  it.  To  get  the 
wo  planes  on  which  r  and  rr  act  in  relative  position,  it  was  neces- 
ary  to  lay  off  r'  =  ORi  and  not  OR'. 

The  obliquity  of  the  stress  for  both  planes  of  rupture  is  <p\ 
ence  IS,  in  its  revolved  position,  is  parallel  to  one  plane  of 
upture  and  IT  to  the  other,  f 

Since  RI  is  often  near  /,  in  fact,  at  /  when  i  =  o,  draw  at  RI 
ccurately  a  perpendicular  to  RiJ  and  mark  the  upper  end  V\ 

*  This  can  be  verified  geometrically.     Thus  regarding,  for  the  nonce,  the 

T 

.rcle  as  having  a  unit  radius,  we  have  in  radians  RIRi  =  -  (R  J  -f  R}  J) . 
ut  this  is  equal  to  the  obtuse  angle  between  the  vertical  plane  and  free  surface, 

rthis  equals  (j  +  «)  =  \  (Ri  J  +  &  -0  +  \  (RJ-R'D  =\  (RJ  +  RiJ). 
t  In  the  general  discussion  at  the  end  of  Art.  58,  OR  and  OR',  Figs.  29  and 
2,  are  any  two  conjugate  stresses  having  the  common  obliquity  FOR  = 
'OR',  which  can  vary  from  o  to  <p.  In  Fig.  36,  the  construction  gives,  as 
2fore,  the  ratio  OR' /OR  for  any  value  of  ROJ  from  o  to  <p.  When  ROJ  =  <?, 
R  =  OR'  =  OS  *=  OT  =  unit  stress  on  either  plane  of  rupture.  The  dis- 
jssion  of  Art.  63  brings  out  the  additional  fact  that  75  and  IT,  in  their 
:volved  positions,  are  parallel  respectively  to  the  two  planes  of  rupture. 


100     NON-COHESIVE  EARTH.   ANALYTICAL  METHODS 

then  measure  the  angles  /3  =  RJT,  y  =  SIV,  made  by  the 
planes  of  rupture  with  the  vertical  R\IV  and  lay  them  off  in 
upper  Fig.  36,  in  true  position,  if  desired.  These  angles  can 
be  found  very  expeditiously  and  accurately  (using  a  large  circle 
of  any  convenient  size)  by  this  diagram,  except  when  i  is  near 
<p,  when  the  intersection  RI  of  OR\  with  the  arc  is  at  too  acute 
an  angle  to  be  reliable.  Since  the  figure  can  be  supposed  to 
refer  to  any  x,  the  plane  of  rupture  remains  at  the  same  inclina- 
tion and  is  continuous  to  the  surface.* 

When  i  =  o,  IV  is  a  vertical  tangent  to  the  circle  at  /  and 
the  planes  IR,  IRi,  IS,  IT,  have  their  true  directions.  Angle 

VIS  =  -  OPS  =  -  (90  —  <p)  .*.    the     planes    of     rupture    lie 

on  either  side  of  the  vertical  and  ^bisect  the  angle  between  the 
vertical  and  the  line  of  natural  slope. 

The  diagram  labors  under  the  inconvenience  of  not  giving 
the  conjugate  stresses  or  the  planes  considered  (except  for 
i  =  o)  in  true  direction,  but  it  is  nevertheless  a  valuable  aid  in 
giving  at  once  the  magnitudes  of  the  conjugate  stresses  and  the 
inclinations  of  the  planes  of  rupture,  and  it  is  especially  valuable, 
in  a  modified  form,  to  coherent  earth. 

NOTE.  After  the  circle  of  Fig.  36  is  drawn  for  the  given  depth  x,  let  it 
be  redrawn  with  RiJ  horizontal.  Then  at  the  new  position  of  0  erect  a  per- 
pendicular to  the  new  position  of  OJ  and  lay  off  on  it  from  O  a  length  equal 
to  the  semi-major  axis  a  =  OJ.  The  semi-minor  axis  01,  in  the  revolved 
position,  is  now  in  true  direction.  On  these  two  semi-axes  of  lengths  a  =  OJ, 
b  =  01,  the  ellipse  of  stress  can  be  drawn  in  true  position.  The  ellipse  of 
stress  is  drawn  by  a  different  procedure  in  Art.  60.  After  the  construction 
of  the  ellipse  of  stress  by  either  method,  the  stress  OR'  on  a  plane  AB,  Fig.  33. 
can  be  found  by  drawing  the  triangle  OP'R',  as  explained  in  Art.  60. 

64.  The  diagram,  Fig.  36,  refers  equally  to  passive  thrust, 
provided  we  suppose  OR',  to  some  scale,  to  equal,  r  =  wx  cos  i  = 

*  The  above  figure  is  equivalent  to  Mohr's  "circular  diagram  of  stress" 
applied  to  earth  pressures.  Mohr's  demonstration  is  more  general  and  along 
different  lines  to  that  given  here.  Prof.  O.  H.  Basquin,  in  a  paper  presented, 
Sept.  9,  1912,  before  the  Western  Society  of  Engineers,  gives  Mohr's  theory 
in  detail,  with  many  interesting  applications.  Also  in  the  Journal  of  the 
Franklin  Institute,  October,  1882,  Mohr's  "Graphical  Theory  of  Earth 
Pressure"  is  presented  by  Prof.  Geo.  F.  Swain. 


64,65]  PLANES   OF  RUPTURE  101 

w  h\  then  rf  —  passive  unit  thrust  on  a  vertical  plane  =  OR  = 
OR",  Art.  58.  It  is  now  easily  shown,  exactly  as  in  the  preced- 
ing article,  that  R'U  =  angle  that  plane  on  which  r  (=  OR') 
iacts,  makes  with  the  minor  axis  and  that  R"IJ  =  angle  that 
plane  on  which  r'  (=  OR"}  acts,  or  the  vertical  plane  makes 
with  the  minor  axis.  Hence  we  must  rotate  the  lower  figure 
jabout  /  until  IRff  is  vertical;  then  IRf  will  be  parallel  to  the 
top  slope. 

Let  tf  =  R"IT,  7'  =  R"IS;  hence  as  IR"  is  now  to  be 
considered  as  vertical,  $  and  7'  give  the  angles  made  by  the 
two  planes  of  rupture  with  the  vertical.  In  this  figure  where 
I  is  near  <p,  y'>  90°,  but  still  a  wedge  of  rupture  is  formed,  as  in 
Fig.  3i(c). 

A  simpler  construction  will  now  be  given  for  finding  all  the 
angles,  /?,  7,  /3',  7',  that  the  planes  of  rupture  make  with  the 
vertical. 

65.  Planes  of  Rupture.  For  measurement  of  angles,  con- 
ceive the  circle  in  Fig.  36  to  have  a  unit  radius,  so  that  any 
icentral  angle  as  SPI  is  directly  measured,  in  radians,  by  the 
arc  IS  which  subtends  it.  Let  D  be  the  mid-point  of  arc  R'S 
and  E  the  mid-point  of  arc  SR,  then  by  geometry  we  have, 

0  =  Z  RJT  =  -  RiT  =  -  R'S  .'.0=  Z  SPD, 

2  2 

-  =  z  siv  =  -2  (is  +  //o  =  ~  (is  +  IR')  :.  7  =  z  DPI\ 

whence,  ft  +  7  =  SPI  =  90°  -  <p. 
Also, 

tf  =  Z  R"IT  =  -  TR"  =  -SR  .'.  $  =  Z  SPE 

2  2 

7'  =  Z  R"IS  =  \  (R"J  +  JS)  =  -2  (RJ  +  SJ)  =  JE. 

:.  7'  =  Z  EPJ  :.  0'  +  V  =  90°  +  *>,  and 

0  +  7  +  t?  +  y  =  180°. 

To  make  an  accurate  construction,  the  scale  should  be  large, 
jand  since  in  similar  figures  corresponding  angles  are  unaltered, 
the  following  construction  is  suggested.  After  laying  off  the 


102  NON-COHESIVE  EARTH.      ANALYTICAL  METHODS 

angles  i  and  <p  from  OJ,  Fig.  37,  take  a  length  of  10  inches  in 
dividers  and  find  by  trial  a  point  P  on  OJ,  from  which  as  center, 
10  inches  radius,  a  semi-circle  can  be  drawn,  as  shown  touching 
OS.  Next,  use  the  dotted  semi-circle  (OP  diameter)  to  find 
very  accurately  the  tangent  point  5.  Then  if  the  line  making 
the  angle  i  with  OJ  cuts  the  first  semi-circle  at  R'  and  R,  bisect 


FIG.  37 

arcs  R'S  and  SR  at  D  and  E,  then  as  proved  above,  0  =  SPD, 
7  =  DPI,  &  =  SPE,  y  =  EPJ.  The  advantage  of  the 
radius  10  inches  is  that  the  angles  can  at  once  be  read  off  from; 
a  table  of  chords.* 

To  derive  formulas  for  0,  etc. ,  we  shall  introduce  an  auxiliary 
angle  c  defined  by  the  relation 

sin  i 

sin  €  =  — ; — . 
sin  <p 

In  the  triangle  OPRf,  we  have, 

sin  i          PR'      PS  sin  i 

• —  = =  —  =  sin <p  .'.  sin  PRO  = =  sin  e. 

sinPR'O       OP       OP  sin? 

Now,  PR'O  =  180°-  [i  +  OPR'}  =  180°-  [i  +  (OPS—20)] 

=  180°-  [i  +  (90°-  <p)  -  2/3] 
.'.  sin  PRfO=  sin  [i  +  90°  —  <p  —  2$  =  sin  e. 
Whence,  i  -f  90°  —  <p  —  2/3  =  € 


o 


9°     -  V       e-»  ,     v 

.  .  0  = ....     (41) 

2  2 

In  the  same  way,  by  considering  the  triangle  OPR,  we  derive, 
0  =  9°.°.±?  _  !±»  (A2) 

2  2 

*  Such  as  is  given  in  Trautwine's  Pocket  Book. 


55]  PLANES    OF   RUPTURE  103 

Having  computed  e  and  then  0  and  0'  from  the  formulas 
ibove,  we  find  7  and  7'  from  the  relations, 


=  90°  -  v,  ff  +  y  =  90° 


*  These  formulas  are  given  by  Resal  in  "Poussee  des  Terres,"  I,  where 
:hey  are  proved  by  a  very  laborious  method.  The  construction  following, 
iimilar  to  Fig.  37,  R6sal  attributes  to  M.  Maurice  d'Ocagne. 


CHAPTER  IV 

DESIGN      OF      RETAINING     WALLS     OF     STONE     OR     REINFORCED 

CONCRETE 

66.  In  the  design  of  retaining  walls,  it  is  generally  specified 
that, 

(1)  The  resultant  on  the  base  shall  cut  it  within  its  middle 
third, 

(2)  The  resultant  on  the  base  shall  not  make  an  angle  with 
its  normal  greater  than  the  angle  of  friction  of  masonry  on  earth, 

(3)  The  soil  pressure  at  the  toes  must  not  exceed  the  safe  bearing 
value  of  the  earth  foundation. 

To  these  requirements  is  often  added  a  fourth,  introducing 
a  so-called  factor  of  safety: 

(4)  The  earth  thrust  on  the  wall,  when  multiplied  by  an  as- 
sumed factor  of  safety  and  combined  with  the  weight  of  the  wall, 
shall  pass  through  the  outer  toe  of  the  wall. 

When  the  earth  thrust  is  computed  by  the  method  which 
includes  the  whole  of  the  wall  friction,  the  writer  recommends 
that  only  the  normal  component  of  the  earth  thrust  be  multi- 
plied by  the  factor  of  safety,  the  friction  component  remaining 
unaltered,  for  reasons  given  in  Arts.  14  and  16.  This  allows 
very  materially  for  the  occasional  decrease  of  <pf  from  repeated 
rains  and  vibration.  For  stepped  walls,  with  no  vibration, 
possibly  both  components  should  be  multiplied  by  the  factor 
of  safety.  For  walls  6  ft.  high,  a  factor  of  safety  of  3.5  may  be 
used;  for  walls  10  to  20  ft.  high,  a  factor  of  3  is  recommended, 
which  may  be  gradually  decreased  perhaps  to  2.5  for  walls  50  ft. 
high  and  upward,  especially  those  with  high  surcharges,  since 
the  cohesive  forces  have  doubtless  come  into  play  before  the 
embankment  is  finished,  thus  materially  decreasing  the  earth 
thrust. 

104 


66,  67]  TENTATIVE    METHOD    OF   DESIGN  105 

After  a  thickness  of  the  base  has  been  found  to  satisfy  (4), 
then  it  must  next  be  ascertained  if  (i)  is  satisfied,  by  combining 
the  earth  thrust  (not  multiplied  by  any  factor)  with  the  weight 
of  the  wall  and  noting  where  the  resultant  cuts  the  base.  If 
it  passes  outside  its  middle  third,  the  thickness  must  be  increased 
until  requirement  (i)  is  satisfied.  Then  it  may  be  ascertained 
if  (2)  and  (3)  are  satisfied,  for  otherwise  the  thickness  will  have 
to  be  still  further  increased.  If  the  thrust  is  computed  by  Ran- 
kine's  formula,  then  this  thrust,  acting  parallel  to  the  surface, 
is  to  be  resolved  into  the  two  components  and  each  multiplied 
by  the  factor  of  safety.  Generally,  when  Rankine's  thrust  is 
used,  requirements  (i),  (2)  and  (3)  are  alone  specified. 

In  an  exceptional  case,  when  the  foundation  is  rock  or  very 
firm  and  the  right  of  way  limited,  the  resultant  on  the  base  may 
be  allowed  to  pass  a  little  outside  the  middle  third,  if  require- 
ments (2)  and  (3)  are  satisfied.  The  whole  base  will  not  then 
be  in  bearing  and  the  leaning  of  the  wall  at  first  will  be  greater 
than  if  (i)  is  satisfied;  but  this  moving  over  of  the  top  of  the 
wall  should  not  increase  with  time  if  the  foundation  is  rock  or 
firmly  cemented  gravel. 

In  connection  with  requirement  (2),  see,  in  addition,  the 
factor  of  safety  against  sliding  discussed  in  Art.  16.  It  is  often 
impracticable,  when  the  base  is  horizontal,  to  make  this  factor 
as  large  as  2,  but  it  should  be  of  course  greater  than  i  and  as 
near  2  as  is  practicable.  Preferably,  the  base  should  be  inclined. 

As  to  the  coefficients  of  friction  and  safe  soil  pressures,  see 
the  tables  of  Art.  8.  As  it  is  very  desirable  to  have  the  resultant 
on  the  base,  not  only  cut  it  witkin  its  middle  third,  but  also 
pass  near  its  center,  to  better  distribute  the  pressure,  a  base  slab 
or  foundation  course  of  masonry,  projecting  beyond  the  outer 
and  inner  faces  of  the  upper  part  of  the  wall,  is  strongly  recom- 
mended. 

Such  a  base  slab  should  effectually  prevent  that  increased 
leaning  with  time,  sometimes  observed,  which  is  doubtless  due 
to  the  very  imperfect  elasticity  of  the  earth,  causing  a  permanent 
set  at  each  increase  of  the  thrust  due  to  rains  or  vibration. 

67.  Tentative  Method  of  Design.     In  Art.  35,  Ex.  5,  Fig.  16, 


106  DESIGN   OF   RETAINING  WALLS 

a  wall  is  designed  by  a  tentative  process,  the  requirement  being 
that  the  resultant  on  the  base  shall  cut  it  one-third  its  thickness 
from  the  outer  toe.  If  no  base  slab  is  used,  it  will  be  more 
satisfactory  to  have  the  resultant  pass  nearer  the  center  of  the 
base — the  amount  to  depend  upon  the  state  of  the  soil  of  the 
foundation,  the  more  uncertain  the  foundation,  the  greater  the 
thickness.  For  a  rock  foundation,  the  thickness  computed  above 
is  sufficient,  but  for  a  soft  soil  it  is  inadequate.  It  is  seen  then 
that  no  hard  and  fast  rule  can  be  laid  down  in  this  case,  but 
that  after  the  thickness  has  been  found  for  minimum  require- 
ments, corresponding  say  to  a  rock  foundation,  the  engineer 
must  use  his  best  judgment  as  to  increasing  the  thickness  of  the 
base  any  farther. 

The  trial  and  error  method  illustrated  in  the  example  cited 
is  of  special  value  when  various  batters  or  thicknesses  at  top 
have  to  be  tried  to  get  a  satisfactory  wall.  Here,  the  top  thick- 
ness and  rear  batter  are  assumed  and  the  front  batter  found  by 
the  construction.  If  unsatisfactory,  another  trial  is  made  and 
so  on,  until  both  batters  as  well  as  the  thickness  at  top  are  found 
to  be  satisfactory. 

In  the  analytical,  direct  method  to  be  given  in  the  next  article, 
the  two  batters  are  assumed  and  the  thicknesses  at  top  and  bottom 
computed.  If  the  latter  .are  unsatisfactory,  other  batters 
must  be  assumed  until  finally  a  satisfactory  top  thickness  is 
found. 

To  save  labor,  the  results  of  computations  for  various  usual 
types  of  walls  have  been  given  in  the  following  tables. 

68.  General  Formula  for  Stability  of  Retaining  Walls  against 
Overturning.  Let  Fig.  38  represent  a  wall  of  length  unity 
perpendicular  to  the  plane  of  the  paper  and  heigh c  h,  its  cross- 
section  being  A  B  C  D,  and  let  the  outer  and  inner  faces  and 
diagonal  D  B  make  angles  with  the  vertical  equal  to  /3,  a, 
and  &  respectively;  also  let  W  denote  the  weight  of  the  wall 
and  g  the  horizontal  distance  from  its  line  of  action  to  the 
outer  toe. 

The  earth  thrust  against  the  face  A  B,  Kwh2,is  supposed 
to  act  at  one  third  of  its  height  above  the  base  and  to  make  an 


68] 


GENERAL   FORMULA    FOR    RETAINING    WALLS 


107 


angle  with  the  normal  to  AB  equal  to  X,  the  normal  component 
of  the  thrust  being  Ki  w  h2  (where  K\  =  K  cos  X)  and  the  tan- 
gential component,  acting  down  in  the  direction  BA,  being 
/  KI  iv  h*,  where  /  =  tan  X. 

The  angle  of  friction  of  earth  on  wall  being  ^/,  when  AB 
lies  above  the  limiting  plane,  Art.  30,  the  angle  X  is  generally  equal 
;  to  <p  when  <?'  ^  <p,  so  that/  =  tan  <p,  is  the  coefficient  of  friction 
of  earth  on  wall.  In  the  design  of  walls,  since  <p'  is  not  generally 
known  in  advance,  it  is  usual  to  put  /  =  <f>,  the  angle  of  friction 
of  earth  on  earth.  For  stepped  walls,  this  is  always  the  correct 
value  to  use.  It  is  implied  here  that  the  thrust  will  be  com- 
puted by  the  method  which  includes 
the  wall  friction. 

When  AB  lies  below  the  limiting 
plane,  the  thrust  and  value  of  X  must 
be  found  after  the  method  outlined 
in  Art.  29  or  preferably  by  the  method 
'  of  Art.  60.  When  the  surface  slopes 
at  the  angle  of  repose,  or  i  =  <p,  the 
formulas  of  Art.  6\  are  directly  appli- 
cable and  furnish  the  shortest  solution. 

When  AB  leans  toward  the  earth 
filling  and  makes  a  small  angle  with 
the  vertical — say  not  exceeding  10° 

—the  thrust  is  computed  as  usual  for  <p  =  <p  and  X  =  <f>.  In 
case  the  wall  makes  a  greater  angle  with  the  vertical  than 
10°,  X  may  be  found  from  Art.  62. 

In  the  formulas  below, 

w  =  weight  of  earth  in  pounds  per  cubic  foot, 
w'  =  weight  of  masonry  in  pounds  per  cubic  foot, 
t   =  thickness  of  wall  at  base  in  feet, 
/'  =  thickness  of  wall  at  top  in  feet, 

o-  =  factor  of  safety  or  number  by  which  the  normal  component 
of  the  thrust  is  multiplied,  the  tangential  component  remaining 
unaltered,  so  that  the  resultant  on  the  base  shall  pass  through 
the  outer  toe. 


FIG.  38 


108  DESIGN  OF  RETAINING  WALLS 

On  taking  moments  about  J9,  we  have, 
Wg  -\-fKi  wh2.  t  cos'a  =  a  KI  wh2  (—sec  a  —  t  sin  a), 

where,  /  =  h  (tan  55  +  tan  a). 

Dropping  the  perpendiculars  CI,  BN,  upon  AD,  we  have 
the  moment  of  the  area  A  BCD  about  D  equal  to, 

-CI.DIX-DI  +  CI.IN(DI  +  -  IN] 

2  3  V  2          / 

+  -BN.NA  (DN  +  -NA] 
2  v        3     ; 

h2  2  r  h 

=  —  tan  0.  -  h  tan  /?  +  h2  (tan  %  —  tan  (3)     htan  (3  +  -  (tan  55 
23  2 

—  tan  0)  I  +  -  h2  tan  a\  h  tan  55  +  -  h  tan  a~\ 

2  f  L  3 

Whence,  on  reducing,  etc., 

w'h*  /  x 

Wg  =  ( 3  tan2  S  +  3  tan  2  /aw  a  +  tan2  a  —  tan2  3 } 

6     V  / 

On  substituting  this  value  and  /  =  h  tan  55  +  h  tan  a    in  the 
first  equation  and  dividing  by  h3, 


w' 


w    / 

—  ( 3  /aw2  a>  +  3  tan  %  tan  a  +  tan2  a  —  tan2 

6   v 

+  /  KI  w  (tan  5  cos  a  +  sin  a) 
=  a  KI  w  f  —  sec  a  —  tan  55  sin  a  —  sin  a  tan  a  ] . 

On  simply  combining  terms,   this  equation  eventually  re- 
duces to, 

tW  n 

— -  (f  cos  a  +  a  sin  a)  2  KI  +  tan  a  \  tan  55 
w 


w  r  a-  -i 

—  2  KI     —  sec  a  —  tan  a  (f  cos  a  +  a  sin  a) 
w'  «-3  J 

-  —  (tan2  a  -  tan2  0)      .->  :i      .   ».     (41)' 


Leygue  gives  this  formula  in  "Annales  des  Fonts  et  Chaussees,"  Nov.,  1885. 


CENTER    OF    PRESSURE    ON    BASE  109 

The  formula  is  adapted  to  the  case  where  the  top  of  the  wall 
leans  toward  the  earth,  or  N  falls  to  the  right  of  A,  on  simply 
changing  sin  a  to  (—  sin  a),  tan  a  to  (—  tan  a)  and  then  regard- 
ing a  as  positive. 

As  this  formula  is  independent  of  h,  it  is  true  for  any  value 
of  h.  If  h  is  given,  tan  co  is  found  by  solving  the  quadratic 
equation  (41),  whence  t  =  h  (tan  co  +  tan  a)  for  AB  battered 
and  /  =  h  (tan  co  —  tan  a)  for  overhanging  walls.  Also  /'  = 
h  (tan  co  —  tan  (3)  for  both  cases.  Should  the  value  of  t'  ever 
become  negative,  the  corresponding  result  must  be  rejected,  since 
there  is  then  no  wall  satisfying  the  assumed  conditions. 

69.  Center  of  Pressure  on  Base  and  Resistance  to  Sliding. 

After  the  value  of  /  has  been  computed  from  (41),  it  is  then  in 

order  to  combine  graphically  the  thrust  Kwh2  on  AB   (not 

multiplied  by  any  factor),  making  the  angle  X  with  the  normal 

to  AB,  with  the  weight  W  of  the  wall  to  find  the  true  resultant 

on  the  base.     Suppose  it  cuts  the  base,  a  feet  from  its  center; 

j  then  according  as  a/t  <  1/6,   =  1/6  or   >  1/6,  the  center  of 

;  pressure  on  the  base  lies  in  the  middle  third,  at  the  third  point 

;  or  outside  the  middle  third  of  the  base.     In  any  case,  if  it  should 

be  thought  desirable  to  alter  t  so  that  the  resultant  shall  cut 

the  base  at  any  prescribed  point,  whether  the  third  point  or 

elsewhere,  the  method  used  in   Ex.  5,  Art.  35,  will  furnish  a 

quick  practical  solution. 

i 
Where  the  resultant  is  required  to  cut  the  base  -  /  from  its 

3 

outer  toe,  the  following  formula,  deduced  in  a  similar  manner 
to  (41),  may  be  used: 

tw  ~i 

—  (/  cos  a  +  sin  a)  4X1  +  tan  (3  \  tan  co 
w  -1 

=  —  2K\  [sec  a  —  2  tan  a  (f  cos  a  +  sin  a)  ] 

w 

-\-  tan  ft  (tan  ft  —  tan  a)    ."    . .    ...    ; .    (42) 

The  formula  refers  to  Fig.  38,  where  N  is  to  the  left  of  A. 

The  formula  is  adapted  to  the  case  where  N  falls  to  the  right  of 


110  DESIGN   OF   RETAINING   WALLS 

A,  or  for  an  overhanging  wall,  on  replacing  sin  a  by  (—  sin  a), 
tan  a  by  (—  tan  a). 

If  the  Rankine  thrust  is  to  be  used  in  connection  with  (42), 
requirements  (i),  (2),  (3)  of  Art.  66  are  alone  specified.  Values 
of  KI  and  X  have  been  given  in  Art.  61  for  various  batters  when 
i  =  (p.  For  any  other  case,  to  find  KI  and  X  after  the  Rankine 
method,  compute  the  thrust  on  a  vertical  plane  through  A, 
Fig.  38,  extending  to  the  surface,  by  Rankine's  formula,  Art.  48, 
corresponding  to  NB  =  i  and  w  =  i.  Combine  graphically  this 
thrust,  acting  parallel  to  the  surface,  with  the  weight  of  earth 
acting  to  its  left,  to  find  the  thrust  on  AB.  Its  angle  with  the 
normal  to  AB  —  X  and  its  component  normal  to  AB  =  KI. 
We  enter  the  formula  (42)  with  this  value  of  KI  and  /  =  tan  X. 

The  angle  6  made  by  the  resultant  on  the  base  of  the  wall 
can  be  easily  found  graphically  or  it  can  be  computed  from 
the  formula, 

E  cos  (X  +  «) 

tan  0  = • — 

W  +  E  sin  (X  +  a) 

which  is  easily  derived  by  aid  of  Fig.  7,  Art.  16,  on  noting  that 
tan  6  =  tan  NGU  =  NU/(GH  +  HU).  On  replacing  /  by  X, 
we  have,  NU  =  E  cos  (X  +  «),  GH  =  W,  HU  =  E  sin  (X  +  a). 
The  formula  follows.  For  an  overhanging  wall,  replace  a 
by  (-  a). 

The  factor  of  safety  against  sliding  can  be  found  as  in  Art.  16. 

Finally,  the  soil  pressures  can  be  computed  from  (3)  or  (4), 
Art.  15. 

70.  Application  of  (41)  to  Various  Types  of  Walls.  In  Fig. 
38,  since  t/h  =  (tan  w  +  tan  a),  if  we  assume  h  =  i,  the  value 
of  /  =  DA  corresponding,  equals  the  ratio  of  the  thickness  of 
the  wall  at  the  base  to  h,  for  any  height  of  wall.  Hence  for 
simplicity,  in  the  following  applications  of  (41)  to  various  types 
of  walls,  it  will  be  assumed  that  h  =  i.  The  values  of  /  and  /' 
will  be  computed  for  <p  =  <p'  =  33°  41',  corresponding  to  the 
slope,  1^/2  base  to  i  rise.  The  expression  for  the  area  of  the 
cross-section  ABCD  =  A  =  y£  (t  +  0>  likewise  represents  the 
volume  of  masonry  for  one  unit  length  of  wall.  Assume  <r  =  3, 
corresponding  to  usual  heights  of  walls  (say  10  to  25  feet)  and 


70-72]      APPLICATION  OF  (41)  TO  VARIOUS  TYPES  OF  WALLS        111 

w       5    w       5 

let  /  and  t'  be  computed  for  the  two  ratios,  —  =  -,  —  =  - 

wf      6'  w'      8' 

f or  i  =  o  and  also  for  i  =  <p. 

When  i  =  o,  the  values  of  K  and  KI,  corresponding  to  the 
case  where  the  wall  friction  is  included,  may  be  taken  from 
Art.  23.  These  values  may  be  computed  from  the  formulas 
of  Art.  44  or  by  aid  of  the  graphical  method  of  Art.  21  or  that 
of  Art.  39.  The  same  remarks  apply  to  overhanging  walls  for 
i  =  o,  but  for  i  =  $  the  formulas  of  Art.  44  will  have  to  be  re- 
sorted to. 

When  i  =  <p  and  the  wall  leans  away  from  the  earth  or  is 
vertical,  the  Rankine  method  applies  and  the  values  of  K  and  X 
will  be  taken  from  the  table  of  Art.  61. 

71.  Type  i.     Vertical  Rectangular  Wall.     Fig.  39. 
a  =  o,  /3  =  o,  t  —  tan  5. 

Formula  (41)  reduces  to, 

w  w 


—  2       i        =   —     2 

w  iv 


When*'  =  o,  K\  =  K  cos  <p'  =  K  cos  <p,  is  found  by  aid  of  (21), 
Art.  46, 


COS2  (f> 


2  (i  +  \2  sin  <f>) 


=  0.109. 


On  substituting  in  the  formula  above,  2  KI  =  o.2i8,/  =  tan  (p  — 
2/3,  we  have, 

w  w 

t2  +  —  (0.145)  /  =  0.218  —  ; 
w  w 

whence  for  w/w1  —  5/6,  /  =  0.371;  for  w/w'  =  5/8,  t  —  0.326. 
When  i  —  <?,  by  (23),  Art.  46,  KI  =  l/2  cos2  <f>  =  0.346, 
whence, 

w  w 

**  +  —,  (0-461)  /  =  0.692  •—. 

w  w 

For  w/w'  =  5/6,  /  =  0.592;   for  w/w'  =  5/8,  /  =  0,530. 

72.  Type  2.    Vertical  Back;  Front  Face  Battered  at  2  Inches 
to  the  Foot.     Fig.  40. 


112 


DESIGN   OF  RETAINING  WALLS 


Here,  a  =  o,  tan  ft  =  1/6  /.  ft  =  9°  28',  /  =  tan  w. 


W 

w 


W  I 

=  —,  2  Ki  +  —  tan2  ft. 
3 


W  w  3 

For  I  =  o;  as  before,  KI  =  0.109,7  =2/3;  whence,  for  w/w'= 
5/6,  /  =  0.381;  for  w/vf  =  5/8,  /  =  0.338. 

When  i  =  <p,  KI  =  0.346,  /  =  2/3,  as  in  Art.  71,  .'.  for 
w/w'  =  5/6,  t  =  0.597;  f°r  w/w'  =  5/8,  /  =  0.536. 

73.  Type  3.  Both  Faces  Battered  2  Inches  to  the  Foot. 
Fig.  41. 

We  have,  tan  a  =  tan  ft  =  1/6  .'.  a  =  ft  =  g°  28'. 


/ 

/v 


FIG.  39 


FIG.  40 


t 
FIG.  41 


FIG.  42 


FIG.  43 


t  =  tan  S>  +  tan  a,    t'  =  tan  Z  —  tan  /?,  tan2  a  ==  tan2  ft 
/  =  2/3,  (7  =  3;  (41)  reduces  to, 


[w  —\w 

—  (l.l5ll)     2    Ki  +  .167        toW  W  =   —  -  2  Ki  X  0.822 
«r  J  ?£/ 

For  i  =  o,  by  (14)  and  (17),  Art.  44,  n  =  0.8434,  KI  =  K  cos  <?= 
0.143. 

Cw  -i  w 

(0.329)  —  -  +  0.167    \  tan%  =  0.235  —  . 
w  J  «r 

w        5 

For  —  =  —  -,  tan%=  0.275  •"•  *  =  0.442,  i  -  0.108. 
w        6 

w        5 

For  —  =  —  ,  tan  w  =  0.240  .*.  t  =  0.407,  tr  =  0.073. 
w        8 

For    i  =  ^   by    Art.   61,   ^  =  0.485,  X  =  31°  58'  .*.  /  = 
tan  X  =  0.624. 

Cw  —i  w 

1.076  —  +  0.167      tow  co  =  0.804  —  ;- 
ze;  J  w;' 


74]  TYPE    4  113 

w        5 

Whence,  if  — -  =  — ,  tan  co  =  0.445  •"•  *  =  0.612,  /'  =  0.278. 
w        6 

w        5 
If  —  =  — ,  tan  co  =  0.404  .*.  /  =  0.571,  r  =  0.237. 

W  o 

74.  Type  4.    Front  Face  Vertical;  Rear  Face  Battered  2 
Iches  to  the  Foot.     Fig.  42. 

|8  =  o,  tow  a  =  1/6,   (7  =  3. 

For  i  =  o,  Ki  =  0.143,  as  in  Art.  73  and  (41)  reduces  to  the 

tan2  a 
neral  formula  of  Art.  73  with =  —  0.009,  added  to 

o 

1  e  right  member. 

r         w  -i  w 

.'.  tan2  co  +     0.329  —  +  0.167      tan  5  =  0.235  —  —  0.009. 
«-          w  -J  MT 

w        5 
.'.  for  — •  =  — ,  tan  co  =  0.266  =  t', 

w         6 

/  =  tan  co  +  tan  a  =  0.266  +  0.167  =  °-433- 

w        5 

—  =  — ,  tan  co  =  0.229  =  t  \  t  =  0.229  "i"  0-167  =  0.396. 

w        8 

To  ascertain  if  (i)  of  Art.  66  is  satisfied,  combine  the  true 
rust  on  the  wall,  K  wh2  =  Kw,  making  the  angle  <p  with  the 
i  aer  face,  with  the  weight  of  the  wall,  w'A ,  to  find  the  resultant 

w        5  i 

its  base.     In  this  case,  K  =  0.172  and  for  —  =  — ,  A  =  — 

W  6  2 

+  /')  =  0.349.     In   the   construction,   it  will  suffice   to  put 
5,  w'  =  6,  whence  the  thrust  =  5  X  0.172  =  0.860  and 
ce  weight  of  wall  =  6  X  0.349  =  2.094.     The  resultant  on  the 
.se  will  be  found  to  pass*  outside  the  middle  third,  not  only 
::  w/wr  =  5/6,  but  also  for  w/wr  =  5/8. 
To  find  the  value  of  /  for  which  the  resultant  on  the  base 
ts  it  at  a  distance  (1/3)  t  from  the  outer  toe,  use  formula  (42). 
us  reduces  to, 

w  w 

tan2  co  +  0.470  — -  tan  co  =  0.212  — -, 

w  w 

<:  substituting,  Ki  =  0.143,  0  =  o,/  =  2/3,  tan  a.  =  1/6. 


114  DESIGN   OF  RETAINING   WALLS 

w         5 
Whence,  for  — -  =  — ,  tan  &  =  0.268  =  t' . 

w        6 

/  =  tan  w  +  tan  a  =  0.435. 

w        5 

For  —  =  — ,  *  =  0.247,  t  =  0.414. 
W        8 

When  i  =  <p,  it  will  be  found  by  a  construction  similar  to  the 
above,  for  the  values  of  t  and  t'  derived  below,  that  the  resultant 
cuts  the  base  within  its  middle  third. 

For  i  =  <?,  from  Art.  61,  K\  =  0.485,  X  =  31°  58'  .'./  = 
tan\  —  0.624;  hence  (41)  reduces  to  (see  Art.  73), 

r  w  -i  w 

tan2  w  +     1.076  — -  +  .167    \tanu>=  0.804  —  —  0.009. 

L  w  J  w 

w        5 

—  =  — ,  tan  %  =  0.440  =  /',  t  =  0.607  J 
w       6 

w        5 

—  =  — ,  tan  c5  =  0.399  =  t ,  t  =  0.566. 
w        8 

75.  Overhanging  Wall.  Front  Face  Battered  2  Inches  to 
the  Foot;  Rear  Face  Parallel  to  the  Front  Face.  Fig.  43. 

For  i  =  o,  <p  =  (pf  =  33°  41',  a  =  9°  28',  we  derive  from 
(19)  and  (20),  Art.  44,  n  =  0.7544,  K  =  0.09745,  Ki  =  0.0811; 
also  the  moment  formula  is  found  from  (41),  by  changing  sin  a 
to  (—  sin  «),  tan  a  to  (—  tan  a),  to  be, 


tan2 


[w  ~\ 

—  (/  cos  a  —  <r  sin  a)  iK\  —  tan  a     tan  w 
w    '  — J 

w          r~  & 
=  —  2Ki     —  sec  a  +  tan  a  (f  cos  a  —  a-  sin  a) 

W  I—    2 


w'         L3 

For    Ki  =  0.081,    /  =  2/3,     a  =  3,    a  =  0  =  9°  28',    this 
reduces  to, 

[w  -i  w; 

0.0267  —  —  0.167  \  tan  co  =  0.169  —7- 
w  J  ttr 

Noting  that  /  =  t'  —  tan  u>  —  tan  a,  we  readily  find, 

w        5 

for  —  =  — ,  tan  u>  =  0.453,  t  =  t  =  0.286; 
w         6 

w        5 

for  —  =  — ,  tan  %  =  0.409,  /  =  t'  =  0.242. 
^        8 


75,  76]  TABLES    OF    RESULTS    FOR    VARIOUS    RATIOS  115 

When  i  =  <p,  by  (19)  and  (20),  Art.  44,  n  =  o,  K  =  0.2999, 
Ki  =  0.2495. 

Hence  the  general  formula  above  reduces  to, 

tw  -i  10 

0.083  ~"7  ~~  0.167      taw  s  =  °-S21  —  > 
ur  J  w' 

w        5 

—  =  — ,  tan  o>  =  0.708,  t  =  t'  =  0.541; 
ur        6 

if        5 

—  =  — ,  tan  55  =  0.631,  /  =  /'  =  0.464. 
w         8 

76.  Tables  of  Results  for  Various  Ratios  w/w'.  The  values 
of  /  and  t1  have  been  similarly  made  out  for  w/w'=  2/3,  w/w'  = 
4/5,  and  all  the  results  checked  graphically  by  combining  the 
components  ^Kiwh2,/  KI  wh2,  for  h  =  i  to  find  the  earth  thrust, 
which  was  in  turn  combined  with  the  weight  of  wall  to  find 
the  resultant  on  its  base. 

It  was  found  in  every  instance  to  pass  nearly  or  exactly 
through  the  outer  toe.  The  results  were  plotted  to  a  large 
scale  and  the  values  of  /,  taken  from  the  corresponding  graph, 
for  w/w'  =  0.60,  0.65,  .  .  .  ,  0.85,  recorded  in  the  following  tables. 
The  results  are  correct  to  within  i  or  2  units  in  the  third  decimal 
figure,  three  significant  figures  only  having  being  used  in  the 
computations.  As  illustrated  in  Art.  74  and  explained  more 
generally  in  Art.  69,  the  true  center  of  pressure  on  the  base 
was  found  and  the  ratio  a/t,  computed;  also  the  angle  0  made 
by  the  resultant  on  the  base  with  the  vertical  was  ascertained. 
Only  rough  average  values  of  a/t  and  0  are  recorded  in  the 
tables,  as  being  sufficient  for  the  purpose  in  view.  In  some  of 
the  cases  of  Type  5,  the  resultant  was  found  to  cut  the  base  to 
the  right  of  its  center,  a/t  then  being  marked  negative. 

In  the  table  referring  to  Type  4,  certain  numbers  are  placed 
in  parentheses.  They  refer  to  the  case  where  the  resultant  on 
the  base  was  required  to  meet  it  1/3  its  width  from  the  outer 
toe.  The  other  figures  were  obtained  by  using  the  factor  of 
safety  method.  It  was  only  for  Type  4  that  the  latter  method 
gave  a  value  of  t  too  small  to  satisfy  requirement  (i)  of  Art.  66. 

The  value  of  0  given  should  not  exceed  the  angle  of  friction 


116 


DESIGN   OF  RETAINING  WALLS 


<p"  of  masonry  on  earth  or  sliding  will  occur.  The  factor  of  safety 
against  sliding,  Art.  16,  will  be  at  least  tan  <p" /tan  0,  and,  if 
possible,  this  factor  should  not  be  less  than  2.  By  reference 
to  the  tabular  values  in  Art.  8,  it  is  seen  that  it  is  difficult  to 
ensure  a  factor  of  2,  especially  when  i  =  #,  so  that  generally  the 
base  should  be  so  inclined  that  the  resultant  on  it,  shall  make 
an  angle  with  its  normal  much  less  than  the  probable  angle  of 
friction. 

TYPE  i 

VERTICAL  RECTANGULAR  WALL, 

V  =  33°  4i',  a  =  3 


Tani 

K 

Ki 

A 

w 
w' 

t  =  t' 

A 

a 

~T 

0 

O 

0.1304 

0.1085 

<P 

0.60 
.65 

.321 
.  ^o 

•  321 

.  ^no 

.126 

11° 

.70 

.  ^40 

.^40 

7S 

^SI 

•7CI 

80 

163 

f 

161 

I  IQ 

12° 

•85 

•374 

•374 

2/3 

0.4161 

0.3462 

<f> 

0.60 
6s 

•523 
S^6 

•523 
S^6 

161 

18° 

7O 

SS2 

SS2 

7S 

S68 

S68 

.80 

S8l 

•  S8l 

.161 

20° 

.8s 

.601 

.601 

TYPE  2 

REAR  FACE  VERTICAL;  FRONT  FACE  BATTERED  2  IN.  TO  i  FT. 
*>  =  33°  4i',  ff  =  3 


Tani 

K 

Ki 

A 

W 

u 

t 

i' 

A 

a 

T 

0 

o 

o  1^04 

o  1085 

(p 

o  60 

T.T.-I 

.166 

.2^0 

.6s 

.  ^4"^ 

.176 

? 

.260 

.064 

13° 

7O 

-2C'7 

186 

.270 

7S 

^6^ 

106 

.280 

•  /o 
80 

^74. 

.  2O7 

.2QO 

OSI 

14° 

8S 

384 

.217 

.  3OO 

2/3 

0.4161 

0.3462 

<f> 

0.60 
6s 

.529 

C4.-2 

.362 

^76 

•445 
.460 

osi 

20° 

7O 

5?^8 

-7QT 

•  47S 

7S 

S74 

4O7 

•  4QO 

80 

•  O/^f 
S8q 

422 

•  SOS 

O4Q 

22° 

8s 

•0°V 
6O2 

4.-7C 

•  SIQ 

76] 


TABLES    OF    RESULTS    FOR    VARIOUS    RATIOS 


117 


TYPE  3 
FRONT  AND  REAR  FACES  BATTERED  2  IN.  TO  i  FT. 

v  =  33°  4i',  °  =  3 


Tani 

K 

Ki 

\ 

w 
w' 

/ 

t' 

A 

a 
T 

I 

O 

0.1724 

0.1435 

<P 

0.60 
6s 

.404 

4IO 

.071 
O77 

.238 

2A_A 

T2T 

H'  V 

•  7O 

.418 

08  S 

•  *'¥\ 
251 

•  75 

.4.27 

OQ4 

260 

80 

417 

IO4 

27O 

T  TO 

1  6° 

8s 

AAA 

III 

.^/U 

278 

•  *•  Ly 

.•*/o 

2/3 

0.5716 

0.4849 

3i°58' 

0.60 
•65 
•7° 

•566 
•576 

.587 

•233 
•243 
•  2S4 

.4OO 
.4IO 
.420 

.084 

23°' 

•75 
.80 
.85 

.596 
.606 
.615 

f 
.263 

•273 
.282 

•430 
.440 
.440 

.076 

24° 

TYPE  4 

FRONT  FACE  VERTICAL;  REAR  FACE  BATTERED  2  IN.  TO  i  FT. 
9  =  33°  4i'.  ff  =  3 


Tan* 

K 

Kj 

A 

w 

«/ 

* 

t' 

A 

a 
T 

e 

o 

o  1724. 

O    I4.^S 

(f> 

o  60 

.701 

.224 

.7,07 

6s 

.4OO 

,2H 

.116 

175 

12° 

7o 

.4.OQ 

.242 

.72S 

.75 

.4.18 

.251 

.774 

.80 

•  4-27 

.260 

.7,47, 

169 

n° 

•85 

•434 

.267 

•351 

0 

0.1724 

0.1435 

<P 

0.6O 
6s 

(.412) 
(.416 

(.245) 
(.240) 

(.328) 

(.772) 

1/6 

! 

7O 

(.420 

(.253) 

(.777) 

75 

(.426 

(.250) 

(.747) 

80 

(.432) 

(.265) 

(.748) 

8s 

yto~/ 
(.417) 

(.270) 

V-Ot"/ 

(.154) 

2/3 

0.5716 

0.4849 

3i°58' 

0.60 
6s 

.562 
.571 

•395 
.404 

.478 
.487 

124 

21° 

7o 

.581 

.414 

.407 

.75 

.501 

.424 

•  5O7 

.80 

.6oi 

.474 

.517 

117 

21° 

•  85 

.6lO 

.447 

.576 

118 


DESIGN   OF   RETAINING  WALLS 


TYPE  5 

OVERHANGING  WALLS,  FRONT  FACE  BATTERED  2  IN.  TO  i  FT.,  REAR  FACE 
PARALLEL  TO  FRONT  FACE 

v  =  33°  4i',  a  =  3 


Tani 

K 

Ki 

A 

•w 
w' 

/  =  t' 

A 

a 
t 

9 

0 

0.0974 

O.oSlI 

<f> 

0.60 
.65 

.238 
.246 

.238 
.246 

—  .OQ 

12° 

.70 

.256 

.256 

•  75 

.267 

.267 

.80 
•85 

•279 
.290 

.279 
.290 

-.07 

13° 

2/3 

0.2999 

0.2495 

9 

0.60 
.65 

•453 
.474 

•453 
.474 

+  -OI^ 

18° 

.70 

.497 

.497 

•75 
.80 

.85 

•512 
•530 
•547 

•  512 
•530 

•547 

+  .019 

19* 

It  has  been  before  remarked,  Art.  40,  that  the  case  i  =  <p, 
or  the  surface  sloping  indefinitely  at  the  angle  of  repose  with 
a  corresponding  plane  of  rupture  of  infinite  extent,  parallel  to 
the  surface,  is  never  realized  in  practice.  The  actual  case  of 
the  surcharged  wall  is  shown  in  Fig.  n,  Art.  32,  where  the  plane 
of  rupture  is  finite  and  the  thrust  is  often  much  less  than  for 
the  slope  of  indefinite  extent. 

The  values  of  t  and  6  in  the  tables  for  i  =  <p,  are  thus  in 
excess  for  any  practical  case.  The  case  of  the  ordinary  sur- 
charged wall  of  one  type  only  will  be  especially  considered  in 
Arts.  79,  80. 

No  foundation  course  or  slab  was  assumed  in  computing  the 
tabular  values.  Where  one  is  used,  the  proper  treatment  will 
be  found  in  the  articles  pertaining  to  concrete  walls. 

77.  Comparison  of  Results.  Applications.  A  comparison  of 
volumes  (column  A),  for  the  batters  assumed,  will  show  that 
for  i  =  o  the  types  are  economical  in  the  order, 

3>  5>  2>  4,  i, 
and  for  i  =  <p  (or  tan  i  =  2/3),  in  the  order, 

3>  2»4>  5,  I- 
The  values  of  a/t,  for  the  various  types,  increase,  for  i  =  o, 

in  the  order,  5,  2,  3,  i,  4,  and  for  i  =  <p,  in  the  order,  5,  2,  3,  4,  i. 
The  variation  is  very  great  and  it  shows  that  if  the  walls  were 


,  78J  COMPARISON    OF    RESULTS  119 

signed  so  that  the  resultant  on  the  base  should  cut  it  one- 
iiird  its  width  from  the  outer  toe,  the  variation  of  the  factor 
i  safety  would  be  very  large.     It  is  on  that  account  that  the 
Titer  prefers  the  factor  of  safety  method,  at  least  in  comparing 
fferent  types  of  walls. 

In  connection  with  sliding,  type  i  shows  the  smallest  value 
0  and  type  3  the  largest  value,  the  variation  for  all  the  types 
t  being  very  marked  for  either  case  of  surface  slope. 
In  designing  walls,  the  weights  'w,  w',  of  earth  and  masonry 
Ibs.  per  cu.  ft.,  have  to  be  assumed.     If  we  assume  w  =  100, 
y  use  of  the  values  given  in  Art.  8,  we  find  the  ratio  w/wf,  for: 

Granite  ashlar o. 61 

Mortar  rubble  granite o. 65 

Mortar  rubble  limestone 0.67 

Reinforced  concrete o.  67 

Plain  concrete o .  70 

Sandstone  ashlar 0.71 

Dry  rubble  granite o.  77 

Dry  rubble  limestone o. 80 

Common  brick  masonry o .  80 

Dry  rubble  sandstone 0.91 

Cinder  concrete 0.91 

Dry  rubble  should  only  be  used  for  low  walls,  for  which  the 
abular  thickness  should  be  increased  x/4  to  1/3  perhaps. 

Recalling,  Art.  70,  that  the  tabular  t  is  the  ratio  of  thickness 
o  height  for  any  height  of  wall  and  similarly  for  /',  consider  a 
Jain  concrete  wall  and  assume  w'  =  143,  w  =  100;  required 
he  thickness  at  base  and  top  of  a  retaining  wall  20  ft.  high,  the 
hickness  at  top  lying  between  1.5  and  3  ft.,  for  i  =  o. 

For  Type  2,  thickness  at  base  =  0.353  X  20  =  7.06  ft., 

"  "        2,  "  "    top     =   0.166  X   20  =   3.33  " 

"       "     3,        "         "  base  =  0.418  X  20  =  8.36  " 
"       "     3,        "         "  top    =  0.085  X  20  =  1.70  " 
Type  3  appears  the  most  suitable  and  it  will  be  chosen, 
especially   since   it   is   the    cheapest.      For   other   ratios   than 
•given  in  the  table,  values  of  /  can  be  interpolated.     The  appli- 
cation is  too  obvious  to  require  an  example. 

78.  As  limited  right  of  way  is  often  a  controlling  feature 
in  retaining  wall  construction,  Type  3  is  the  favorite,  especially 


120 


DESIGN   OF   RETAINING   WALLS 


as  it  is  economical  and  the  form  work  for  concrete  is  simpler 
than  for  the  overhanging  wall,  which  is  next  to  it  in  economy. 
On  that  account;  the  following  table  is  presented  for  that  type, 
only  with  the  front  face  battered  i  in.  to  the  foot,  the  rear  face 
being  battered  o,  i,  2,  3,  4,  5  in.  to  the  foot  in  turn.  As  before. 
<p  =  33°  41'  or  tan  <p  =  2/3  and  (7  =  3. 

For  the  case  i  =  o,  tan  a  =  5/12,  the  computed  value  of  t' 
is  negative  and  there  is  no  solution.  If  we  suppose  tf  =  o  with 
the  computed  values  of  t,  we  have  the  case  of  triangular  walls 
corresponding  to  a  safety  factor  very  slightly  greater  than  3. 

It  will  be  observed  that  A  decreases  as  tan  a  increases,  but 
this  saving  in  volume  is  offset  by  an  increase  in  6,  particularly 
when  tan  i  =  2/3  or  i  =  <p.  The  angle  of  friction  of  masonry 
on  wet  clay  has  been  .given  from  11°  to  18°;  for  sand,  dry  clay  and 
gravel,  22°,  27°  and  31°,  respectively.  Evidently  from  the  values 
of  0  given  in  the  table  for  i  =  <p,  the  bases  must  be  tilted  upward. 

TYPE  3 

FRONT  FACE  BATTERED  i  INCH  TO  FOOT,  REAR  FACE  WITH  BATTERS  RANGING 
FROM  o  TO  5  INCHES  TO  THE  FOOT 


Tani 

Tan  a 

K 

Ki 

A. 

w 
w' 

/ 

/' 

A 

a 

T 

0 

O 

O 

0.1304 

0.109 

<f> 

2/3 

4/5 

•339 
.366 

•  256 

.283 

.298 

.  325 

.088 
.082 

12° 
12.6° 

•  •  • 

1/12 
2/12 

0.1520 
O.I7I5 

O.I27 
0.143 

/ 
2/3 

4/5 
2/3 

4./C 

•378 
.402 
.407 
470 

.212 
.236 

•157 
1  80 

•295 
•319 

.282 

.  305 

.127 
.117 
•133 

1^5 

12.6° 

13-5° 
15-5° 
14.  2° 

... 

3/12 

O.I97O 

0.164 



/° 
2/3 
4/5 

•431 
45O 

.098 
.  117 

.264 
.283 

•I63 
.167 

12.8° 
IS-  1° 

... 

4/12 

0.2245 

0.187 

•t/o 

2/3 
4/5 

•453 
.460 

•037 

•°53 

•^ 

' 

.167 

.167 

14° 

15-5° 

5/12 

0.2525 

O.2IO 

t/  o 

2/3 

4/c 

.460 

474 

2/3 

O 

I/I2 

O.4l6l 
0.4922 

0.3462 
0.4119 

33°4i' 
33°  'i  i' 

q/a 

2/3 
4/5 
2/3 

4/5 

•544 
-56i 
•557 
.583 

.461 

.478 
•  391 
.417 

.503 
1  520 

•474 
.  500 

•074 
.084 
.IOO 
•  OQ4 

19-3° 
22° 
21.2° 
22.  6° 

2/!2 

0.5716 

0.4849 

3i°  58' 

/ 
2/3 

4/5 

•576 
.602 

.326 
.352 

•451 
•477 

.115 
.  Ill 

22.1° 

*V7° 

... 

3/12 

4/12 

0.6595 
0-7543 

0.5697 
0  .  6646 

30°  15' 
28°  14' 

/ 
2/3 

4/5 

2/3 

4/5 

•583 
.603 

.583 
598 

.250 
.270 
.167 

.182 

.417 
•437 
•375 
•39° 

.138 
.130 
•153 
.  164 

23.7° 
25-3° 
24.8° 

3i° 

5/12 

0.8566 

0.7696 

26°  03' 

2/3 

4/5 

.582 
•  593 

.082 
.093 

•332 
•343 

-.182 
.184 

26.1° 

27.  Q° 

5/12 

0.8566 

0.7696 

26°  03' 

/ 
2/3 

4/5 

(.663) 
(-609) 

(.103) 
(.109) 

(-353) 
(-359) 

(.167) 
(-167) 

25-*°0 

JZii! 

79] 


SURCHARGED    WALLS 


121 


79-  Surcharged  Walls,  Fig.  44;  Type  2 ;  Back  of  Wall  Vertical, 
Exterior  Face  Battered  2  Inches  to  the  Foot.    The  earth  surface 
extends  from  C,  at  the  angle  of  repose  upward  to  a  horizontal 
surface  /L,  a  vertical  distance  h'  above 
the  top  of  the  wall,  whose  height  is  h. 
Assume  .X  =  <p  =  <?' ;  then  to  find  KI, 
for  a  given  h'/h,  as  */£,  lay  off  h  —  10 
(say)  and  h'  =  5  and  by  the  construc- 
tion of  Fig.  19,  Art.  38,  for  w  =  i, 
find   E  =  K  h2  .'.  K  =  E/h*  and  #1 

«  D 

=  K  cos  <p.  FlG  44 

To  find  the  center  of  pressure,  the 

values  of  c  of  Art.  43  were  plotted,  the  points  connected  with 
a  curve  and  the  values  for  various  ratios  of  h'/h  taken  from 
the  graph,  some  of  them  being  only  roughly  approximate. 
The  interpolated  values  are  enclosed  in  parentheses.  The 
values  of  /  were  found  by  the  use  of  (41)  of  Art.  68  modified, 
using  the  values  of  c  and  K\  as  derived  above. 
SURCHARGED 


h 

» 

h 

Ki 

c 

/ 

h1 

W/W'=  2/3 

w/w'  =  4/5 

o.o 

00 

•346 

•333 

•548 

•589 

O.I 

10.00 

.313 

(-338) 

•532 

•570 

O.2 

5.00 

•293 

(-341) 

.522 

•559 

0-3 

3-33 

.279 

(•344) 

•SIS 

•553 

0.4 

2.50 

.270 

(-347) 

511 

•548 

0.5 

2.00 

.261 

•351 

•507 

•545 

0.6 

1.67 

•253 

•354 

•504 

•541 

0.8 

I-25 

.238 

.360 

•497 

•534 

1.6 

I  .00 

.225 

•364 

.488 

•524 

1.2 

0.83 

.214 

•364 

•479 

•  5i6 

i-5 

0.67 

.202 

•364 

.468 

•504 

2.O 

0.50 

.188 

•364 

•456 

.489 

3-0 

0-33 

.166 

(-363) 

•432 

•465 

5-o 

0.20 

•143 

(.360) 

.408 

•439 

IO.O 

O.IO 

.130 

(-355) 

•385 

.414 

00 

o.oo 

.109 

•  333 

•346 

•374 

Summary  of  results  for  surcharged  wall,  Fig.  44.     <p  =  <?'  = 
<r  =  3,  a  =  o,  tan  0  =  1/6,  h  =  i.     The  width  of  base 
for  any  h  =  (tabular  value  of  t)  X  h. 

The  value  of  0,  or  the  angle  made  by  the  true  resultant 


33°  4i' 


122 


DESIGN   OF  RETAINING  WALLS 


(for  o-  =  i)  on  the  base  with  the  vertical  varies  from  13°  to  14° 
when  h'/h-=  o,  to  20°  to  22°,  when  h'/h  =  <*>. 

80.  Wall  with  Surcharge  Extending  over  the  Top.    Fig.  45. 
Type  2 ;  back  of  wall  vertical;  exterior  face  battered  2  inches  to  the 

foot.  The  earth  surface  extends 
from  Z),  upward  at  the  angle  of 
repose,  to  a  horizontal  surface 
7L,  a  vertical  distance  hf  above 
the  top  D  C  of  the  wall. 

The  values  of  KI  and  c  may 
be  derived  from  those  given  in 
the  preceding  table,  for  the  ratio 
N  M/B  N  of  Fig.  45  corresponds 
to  h1 ' /h  of  Fig.  44,  in  finding  K\ 
and  c.  Let  BN  =  h0\  then  tak- 
ing moments  about  A,  as  in  Art.  68,  we  derive,  after  reduction, 

W  fno\2  W  fno\Z  I 

tan2  S>  +  /  —  2Ki  [  — )    tan<Z=<rc  —  2X1  ( —  J   -\ tan2  0, 

ur          \h  '  w  \k'         3 

the  symbols  5>,  0,  o-,  w  and  w',  having  the  meanings  given  in  Art.68. 

The  weight  of  earth  D  C  N  over  the  wall  was  neglected  in 

rinding  the  resisting  moment,  which  is  on  the  side  of  safety. 

BURIED  WALLS 


FIG.  45 


p 

1 

h' 

i 

h 

w/w'  =  2/3 

w/w'  =  4/5 

h 

w/w'  =  2/3 

w/w'  =  4/5 

0 

•  346 

•374 

2.6 

.830 

.942 

O.I 

.406 

•450 

2.8 

.838 

•950 

0.2 

.452 

•494 

3- 

.845 

•958 

o-3 

.490 

•540 

3-5 

.860 

•974 

0.4 

.520 

.580 

4- 

872 

•989 

0.5 

.556 

.617 

4-5 

.80; 

.004 

0.6 

.586 

-652 

5- 

.890 

.017 

0.7 

.614 

.680 

5-5 

.895 

•  025 

0.8 

.636 

.710 

6. 

.900 

•  032 

0.9 

.658 

•733 

6-5 

.899 

.038 

I.O 

.675 

•753 

7- 

.902 

.041 

1.2 

.708 

.792 

7-5 

.904 

•043 

1.4 

•735 

.830 

8. 

.906 

.047 

1.6 

•757 

.862 

9 

.909 

•053 

1.8 

2.0 

.776 
.792 

.890 
.910 

10 

15 

.912 
.927 

.080 

2.2 

.807 

.924 

20 

•943 

.104 

2.4 

.820 

•934 

00 

I.OI2 

.171 

WALL  WITH  SURCHARGE  EXTENDING  OVER  THE  TOP         123 

Summary  of  results  for  wall  with  surcharge  extending  over  the 
/«>.  <p  =  <p'  =  33°  41',  a-  =  3,  a  =  o,  tan  ft  =  1/6,  /  =  tan  <p 
•-2/3,  h  =  i.  The  width  of  base  for  any  h  =  (tabular  value 
<»/)  X  h. 

In  the  computations,  a  trial  /'  was  assumed,  then  CN  and 
.'V  computed.  For  the  ratio,  NM/BN  (corresponding  to 
n'h  of  Fig.  44)  the  values  of  KI  and  c  were  interpolated  from 

!>  table  of  Art.  79.     Inserting  these  values  in  the  formula, 
a  co  =  t  (for  h  =  i)  is  computed  and  then  t'  =  t  —  0.167. 
!!  this  value  of  /'  does  not  agree  with  the  one  assumed,  a  new 
,al  t'  must  be  taken  and  the  work  repeated.     Generally  one  or 
o  trials  sufficed. 
The  true  resultant  on  the  base  makes  an  angie  with  the 

[Ttical  which  varied  from  13°  to  14°  for  h' '/h  =  o  to  23°  to  26° 
•  li' lli  =  oo .     Provision  must  be  made  for  these  large  in- 
c  nations. 

EXAMPLE.     In   Fig  45,   suppose  h  =  BC  =  20  ft.,   h'  =  CM  =  120  //. 

.'V/h  =  6  .'.  the  width  AB  of  wall,  for  the  assumptions  above,  and  w/w'  =  ^ 
uld  be  20  X  0.9  =  i8ft. 

The  table  of  thickness  for  the  buried  walls  was  made  out 
oressly  to  show  the  greatly  increased  values  of  t  required  when 
.2  earth  is  allowed  to  extend  over  the  top  of  the  wall.  If 
'11s  have  been  designed  for  the  previous  case  and  the  earth 
i:  carelessly  dumped  over  them,  failure  will  almost  certainly 

An  inspection  of  the  last  two  tables  will  show  that  the  values 
<:  /  vary  very  much  with  the  height  of  the  surcharge.  When 
.;'/?  is  large,  the  result  approaches  that  for  i  =  <p,  which  can 
1:  substituted  for  it,  the  solution  for  i  =  <p  being  very  simple. 

When  heavy  loads,  as  locomotives,  rest  on  the  level  portion  of 
:s  surcharge,  the  weight  must  be  replaced  by  an  equal  weight 
j  earth  and  the  method  of  Art.  32  applied  or  when  h' '/h  is 
-  *ge,  the  solution  for  i  =  <p  may  be  sufficiently  near,  remember- 
:  *  that  in  the  latter  case,  the  thrust  is  applied  at  %  the  height, 
"'icreas  in  the  former  case  the  point  of  application  is  higher. 

In  the  computations  pertaining  to  the  previous  tables,  it  is 


124  DESIGN   OF  RETAINING  WALLS 

well  to  recall  that  for  i  =  <p,  the  Rankine  method  was  used  in 
rinding  the  thrust  on  other  than  overhanging  walls.  It  is  exact 
for  such  cases  and  furnishes  the  values  of  KI  and  X  needed  for 
use  in  (41).  From  the  examples  worked  in  Art.  35,  it  is  seen 
to  be  very  inexact  when  i  is  small  or  zero,  for  walls  having  little 
or  no  batter  on  the  interior  face;  hence  the  method  including  the 
wall-friction,  was  used  throughout  for  level-topped  fining.  How- 
ever, as  noted  in  Art.  35,  the  two  methods  do  not  differ  much 
in  their  results  when  the  interior  batter  is  large — say  3  to  5  in. 
to  i  ft.,  so  that  the  Rankine  method  can  be  used  for  such  batters 
where  desirable.  It  is  fortunate  that  it  is  so,  for  the  Rankine 
method  is  especially  convenient  where  the  filling  is  subjected  to 
heavy  loads  and  it  will  be  freely  used  in  the  following  articles 
on  the  design  of  reinforced  concrete  walls. 

Remark.  It  is  important  to  use  a  proper  material  for  the 
filling  of  retaining  walls.  A  granular  material  as  sand,  gravel  or 
rip-rap  is  good,  but  as  the  amount  of  clay  increases,  the  physical 
characteristics  are  more  variable,  especially  at  the  time  of  heavy 
rains,  which  may  materially  alter  the  coefficients  of  friction 
and  cohesion  unless  the  drainage  is  efficient. 

It  is  inadvisable  to  use  clay  in  a  nearly  pure  state  as  a  filling, 
especially  the  kind  of  clays  that  swell  on  exposure,  sometimes 
with  enormous  force. 

Clay  in  bank  often  stands  at  a  steep  slope,  even  for  years, 
and  then  gives  way.  Under  the  influence  of  the  sun  and  the 
weather  generally,  it  has  been  known  to  run,  somewhat  like  a 
viscous  substance.  In  fact  its  properties  are  so  variable  that 
each  clay  bank  has  to  be  studied  by  itself.  For  a  foundation 
course,  protected  from  the  atmosphere,  it  is  generally  reliable 
when  sufficiently  stiff;  but  when  excavated  and  dumped  behind 
a  retaining  wall,  it  is  difficult  to  say  what  its  pressure  may 
become  in  the  course  of  time. 

81.  Reinforced  Concrete  Retaining  Walls.  In  the  Appendix 
will  be  found  a  discussion  of  reinforced  concrete  beams  or  slabs, 
such  as  are  met  with  in  the  design  of  retaining  walls,  where  the 
faces  are  often  inclined  to  each  other  and  sometimes,  more  than 
one  set  or  layer  of  rods  is  used.  In  the  case  of  the  wedge- 


REINFORCED    CONCRETE    RETAINING    WALLS  12o 

ped  slabs,  a  certain  assumption  was  necessary  to  obtain 
orkable  formulas.  The  results  are  on  the  side  of  safety  and 
lis  feature  should  appeal  to  practical  men. 

The  usual  formulas  and  diagrams  for  prismatic  beams,  are  ob- 
tined  as  special  cases  of  the  general  solution.  Certain  practical 
ata  are  given,  near  the  end  of  the  Appendix,  relating  to  the 
est  practice  of  the  present  day,  which  should  be  carefully  read. 
I  the  reader  has  already  studied  the  subject  of  reinforced  con- 
rete  beams  in  existing  text  books,  he  will  be  all  the  better 
repared  to  follow  the  reasoning  and  make  proper  applications 
f  the  formulas  and  methods  proposed. 

The  concrete  generally  used  in  retaining  wall  construction 
>  a  mixture,  by  volume,  of  one  part  Portland  cement  to  two 
•arts  sand  or  fine  aggregate  to  four  parts  coarse  aggregate,  or 
s  usually  abbreviated,  a  i  :  2  :  4  concrete. 

To  save  repetition,  in  all  the  examples  that  follow,  the  ratio 
>f  the  modulus  of  elasticity  of  steel  to  that  of  the  concrete,  will 
;>e  taken  as  n  =  15  and  the  following  working  stresses  will  be 
ised: 

tensile  stress  in  steel /,  =  16,000  Ib.  sq.  in. 

lending  stress  in  concrete fc  =  650  Ib.  sq  in. 

>hear  as  a  measure  of  diagonal  tension  in  a  beam 

without  web  reinforcement v  =  40  Ib.  sq.  in. 

Concrete  in  shear  not  combined  with  tension  or 

compression-punching  shear 120  Ib.  sq.  in. 

3ond  stress  on  plain  round  or  square  rods u  =  80  Ib.  sq.  in. 

3ond  stress  on  plain  round  or  square  rods  with 

hooked  ends,  bent  180°  to  a  radius  of  3  diameters 

and  with  a  short  length  beyond  the  bend u  =  100  Ib.  sq.  in. 

Bond  stress  on  deformed  bars,  varying  with  the 

form u  =  So  to  150  Ib.  sq.  in. 

To  secure  sufficient  bond  resistance,  plain  bars  shall  be 
embedded  in  the  concrete  50  diameters  and  deformed  bars  or 
plain  bars  with  hooked  ends  (with  an  assumed  u  =  100  Ib.  sq. 
in.  for  either)  40  diameters. 

The  soil  pressure  allowed  will  be 5,ooo  Ib.  sq.  ft. 

It  is  often  the  case  that  the  bearing  power  of  the  soil  deter- 
mines the  length  of  base  of  the  wall.  The  safe  bearing  capacity 


126  DESIGN   OF  RETAINING   WALLS 

of  dry  sand  or  clay  mixed,  with  sand,  is  from  4000  to  6000  Ib. 
sq.  ft.  and  for  more  compacted  soils  or  those  with  an  admixture 
of  gravel,  over  8000  Ib.  sq.  ft.;  but  for  soft  clay,  2000-4000  and 
for  quicksands,  alluvial  soils,  1000-2000  Ib.  sq.  ft.  only  is  allowed. 

It  is  very  necessary  to  thoroughly  drain  the  filling,  which 
can  be  effected  by  drains  placed  just  back  of  the  wall  and  on 
top  of  the  footing  and  by  "weep  holes"  through  the  wall  at 
intervals  of  about  25  feet.  The  weep  holes  may  be  from  3" 
to  4"  diameter  and  must  extend  through  the  wall,  just  above 
the  ground  surface,  where  a  longitudinal  drain  should  be  con- 
structed to  carry  off  the  water. 

Plain  concrete  walls  should  have  vertical  expansion  joints 
40  ft.  apart  at  most.  The  stresses  due  to  shrinkage  and  tem- 
perature changes  in  reinforced  walls,  are  usually  provided  for 
by  the  steel,  in  amount,  at  least  ^3  of  i%  of  the  cross-section 
of  the  wall  and  well  distributed  near  the  exposed  surface  of 
the  concrete. 

The  coefficient  of  friction/'  of  masonry  on  dry  clay  is  given 
as  0.5  to  0.6  and  on  wet  clay,  as  only  0.2  to  0.33,  the  latter 
figure  referring  to  moist  clay.  On  dry  earth,  it  is  about 
%  to  %  and  on  dry  sand  or  gravel  ^3  to  ^. 

Sliding  is  to  be  feared  in  light  walls  and  the  resistance  of  the 
wall  to  sliding  should  be  the  first  thing  tested.  The  weight  of 
wall  and  filling  just  over  it,  multiplied  by  the  coefficient  of 
friction,  should  be  greater  than  the  horizontal  earth  thrust. 

The  coefficient  of  friction  /'  of  wall  on  earth  will  generally 
be  taken  at  0.5  in  what  follows. 

In  all  the  retaining  walls  that  will  be  examined,  the  following 
values  are  specified: 

w  =  weight  of  earth  per  cu.  ft.  =  100  Ibs., 
w'  =  weight  of  concrete  per  cu.  ft.  =  150  Ibs., 
angle   of   repose   of   earth   <p  =  33°  41',   corresponding   to   the 
natural  slope  2  on  3  and  to  a  coefficient  of  friction/  =  tan  <p  =  %> 
The  surcharge  will  be  taken  at  800  Ib.  sq.  ft.  on  the  level  sur- 
face of  the  earth,  corresponding  to  an  additional  load  of  earth 
8  ft.  high.     The  design   of   the  following  walls  is  in  reality  a 
partly  tentative  process,  requiring  a  little  preliminary  testing 


81,  82]     DESIGN  OF  A  TRAPEZOIDAL  WALL  WITH  BASE  SLAB          127 


of  assumed  dimensions.    To   save   repetition,   only   the  final 
design  will  generally  be  given. 

The  width  at  the  top  has  been  taken  at  2  feet.  Perhaps 
1.5  ft.  or  less  would  suffice  for  highways  near  the  wall  and 
something  over  2  feet  would  be  more  desirable  where  railway 
tracks  are  placed  parallel  to  the  wall.  In  the  first  example 
below,  railway  tracks  will  be  assumed, 
and  since  the  trains  can  occupy  either 
or  all  of  the  parallel  tracks,  the  sur- 
charge should  be  taken  as  extending 
to  a  point  or  points  that  will  sub- 
ject the  wall  or  the  soil  to  the  most 
trying  conditions. 

82.  Design  of  a  Trapezoidal  Wall 
with  Base  Slab.  In  Fig.  46  is  shown 
the  first  wall  that  will  be  examined, 
the  dimensions  being  marked  on  the 
figure.  Let  us  first  examine  the  part 
A  B  CD  above  the  footing. 

Conceive  the  surcharge  as  extend- 
ing up  to  a  vertical  plane  through  B. 
The  height  of  this  plane  =  h  =  20',  height  of  surcharge  = 
h'  =  8';  hence  by  eq.  (27),  Art.  51,  the  earth  thrust  on  this 
plane,  acting  horizontally  to  the  left,  as  given  by  Rankine's 
formula,  for  a  foot  length  of  wall  is, 


FIG.  46 


E  =    2  tan* 


(45°  -  —  )  w  [  (h  +  h'Y  ~ 


=  0.143  X  100  (282  -  82)  =  10,296  Ibs. 
and  it  acts  above  B}  a  distance,  Art.  51,  eq.  (28), 
h' 


0  = 


\  h 

i-  = 

'' 


ft- 


h  +  *K'  3         ^  9  '  3 

.'.  EC  =  10,296  X  8.15  =  83,900  ft.-lbs. 
In  finding  the  center  of  gravity  of  any  area,  always  divide 
it  up  into  right  triangles  and  rectangles.     One  side  of  each  triangle 
is  vertical  and  a  vertical  line  through  its  center  of  gravity,  cuts 
the  horizontal  base,  ]^  its  length  from  the  vertical  side.     To 


128 


DESIGN   OF  RETAINING  WALLS 


find  how  far  the  vertical  through  the  center  of  gravity  of  the 
combined  weight  of  wall  above  AB  and  of  the  earth  over  BC, 
acts  from  A,  take  moments  about  A. 


Weight,  Pounds 

Arm,  Feet 

Moment,  Foot-Pounds 

25  X  150 

40  X  150 
75  X  150 
75  X  100 

2-5  +  1       = 
4-5  +  2-5   = 
4-5+5       = 

3-5 
7.0 

9-5 

6,250 
2I,OOO 
78,750 
71,250 

28,500 

177,250 

The  line  of  action  of  the  resultant  weight  is  thus  to  the  right 
of  A  ,  a  distance, 

I77,25O/28,5OO  =   6.22  ft. 

A  good  check  is  afforded  by  taking  moments  about  some  other 
point  than  A. 

We  next  combine  the  weight  above  with  E,  acting  8.15  ft. 
above  B.  Suppose  the  resultant  cuts  the  plane  AB  at  a  point 
distant  x  from  A.  Taking  moments  about  this  point;  the 
moment  of  the  resultant  is  zero  .'.  the  algebraic  sum  of  the 
moments  of  its  components  is  equal  to  zero. 

.'.  28,500  (6.22  —  x)  =  EC  =  83,900 
.*.  x  =  3.27  ft. 

Thus  the  base  being  12  ft.  in  length,  the  resultant  cuts  it, 
0.73  ft.  outside  of  the  middle  third  and  6  —  x  =  2.73  ft.  from 
the  center  of  the  base.  The  vertical  component  of  this  re- 
sultant =  weight  =  28,500  Ibs.,  hence  the  unit  pressures  at 
A  and  B,  are,  Art.  15,  eq.  (3), 


*  w 

l 


12 


:.  stress  at  A  =    5,605  Ib.  sq.  ft.  =     39  Ib.  sq.  in. 
stress  at  B  —  —  855  Ib.  sq.  ft.  =  —  6  Ib.  sq.  in. 
and  about  20  in.  of  the  base  AB  is  under  tension,  varying  from 
o  to  6  Ib.  per  sq.  in. 

The  concrete  is  perfectly  able  to  withstand  this  tension,  but 
for  a  possible  increase  of  the  thrust  from  heavy  rains  and  vibra- 
tion, place  K"  D  steel  bars,  say  3"  from  B  C  and  20"  center 


2,  83]  DESIGN    OF    BASE    SLAB  129 

o  center.  The  bars,  placed  parallel  to  B  C,  may  extend  2 .5  ft. 
•clow  AB  to  develop  sufficient  bond  strength  (50  diameters  = 
5")  and  say,  4  feet  above  AB.  The  total  tension  on  AB  for 
in.  length  of  wall  =  YL  X  6  X  20  =  60  and  on  20  in.  length 
'f  wall  (the  spacing  of  the  bars)  =  1200  Ib.  The  allowable 
tress  on  YI"  D  bar  =  %  X  16,000  =  4000  Ibs.,  so  that  the 
ods  are  seen  to  be  ample  to  take  all  the  tension  without  enter- 
ng  into  a  precise  computation.  If  AB  is  diminished  in  length 
he  amount  of  tension  increases  and  the  wall  approaches  the 
f  form,  which  is  considered  fully  further  on. 

83.  Design  of  Base  Slab.    We  next  consider  the  stability 
»f  the  entire  wall  when  the  surcharge  extends  to  N.    Now, 
=  24,  h'  =8,  .'.  as  above, 

E'  =  0.143  X  I0°  (32'  -  82)  =  13*728  Ibs. 
,nd  the  horizontal  thrust  E'  acts  above  H, 


— 


9.60  ft. 


24  +  i6/  3 
E'c'  =  131,800  ft.-lbs. 
}'  is  the  earth  thrust  on  the  plane  HN  and  E'c'  is  its  moment 
-bout  any  point  in  LH. 

The  weight  of  the  whole  wall,  with  that  of  the  earth  over 
t,  is  next  to  be  found,  as  well  as  its  line  of  action. 

'he  weight  of  footing  ALH  =  (4  X  17  —  K  X  4  X  3)  150.  .  =    9,300  Ibs. 

Veight  of  earth  over  BM  =  20  X  100 =    2,000    " 

Veight  of  A  BCD  and  earth  over  it =  28,500    " 

Total  weight  of  wall  and  earth  over  it =  39,800    " 

To  utilize  moments  previously  found,  moments  must  still  be 
aken  about  A. 

Divide  the  footing  cross-section  into  the  rectangle  A  H  and 
he  trapezoid  AL,  which  last,  further  subdivide  into  the  rect- 
mgle  4'  X  i'  and  the  remaining  triangle. 

Taking  moments  about  A , 

Jl  /  A\  ~~| 

13*  X  4  X  —  -  fiX4X2+#X3X4X— )     150.  =    48,300  ft.-lbs. 

vlt.  earth  wt.  over  BM  =  20  X  100  X  12.5 =    25,000       " 

foment  previously  found =  177,250       " 

250,550       " 


130  DESIGN   OF   RETAINING   WALLS 

Thus  the  total  weight  of  the  entire  wall  and  that  of  the  earth 
directly  over  it,  is  w'  =  39,800  Ibs.  and  its  moment  about  A  is, 
250,550  ft. -Ibs.  Its  line  of  action  is  thus  to  the  right  of  A, 
a  distance, 

25o?55o  ., 

-  =  6.29  ft, 
39,800 

and  to  the  right  of  L,  a  distance,  4  +  6.29  =  10.29  ft. 

Let  the  resultant  of  Ef  and  W'  cut  the  base  LH  at  a  point 
x'  ft.  from  L.  Taking  moments  about  this  point, 

W  (10.29  -  x'}  =  E'c' 
or,  39,800  (10.29  —  x')  =  131,800, 

.'.  x'  =  6.98  ft. 

The  resultant  thus  cuts  the  base,  8.5  —  6.98  =  1.52  ft.  from 
the  center  of  LH. 

The  unit  stress  on  the  earth  foundation  at  L  and  H  is, 

39,800  /         6  X  i.52\ 
P  =    \*  '-    J  ==  234o  (i  =*=  0.537) 

.'.  soil  reaction  at  L   =  3600  Ib.  sq.  ft.  =    25  Ib.  sq.  in. 
soil  reaction  at  H  =  1080  Ib.  sq.  ft.  =  7.5  Ib.  sq.  in. 
which  is  safe  for  ordinary  (not  soft)  clay  soils  or  for  soils  of 
ordinary  clay  mixed  with  dry  sand.     The  soil  pressures  at  L 
and  H  may  be  increased  by  supposing  the  surcharge  to  extend 
up  to  C;    but,  as  in  this  instance,  the  increase  will  most  prob- 
ably not  be  enough  to  make  the  maximum  soil  pressure  over 
5000  Ib.  per  sq.  ft.,  which  is  allowed,  the  possible  increase  was 
not  computed. 

It  is  well  to  lay  off  the  upward  soil  pressures  at  L  and  H, 
3600  and  1080  Ib.  sq.  ft.  respectively,  connect  with  a  straight 
line  and  measure  to  scale  the  soil  pressure  vertically  under  A. 
It  is  found  to  be  3000  Ib.  sq.  ft.  The  total  soil  pressure  on  the 
base  of  the  toe  AL  is  thus  >^  (3600  +  3000)  X  4  =  13,200  Ibs. 
(for  one  foot  length  of  wall)  and  it  acts  through  the  center 
of  gravity  of  the  trapezoid  or  2.1  ft.  to  the  left  of  A. 

In  finding  this  distance,  we  can  proceed  either  graphically 
(see  Art.  34)  or  analytically.  In  the  latter  case,  divide  the 
trapezoid,  having  the  vertical  sides  3600  and  3000,  into  a  rect- 


>]  DESIGN    OF    BASE    SLAB  131 

angle  of  area  4  X  3000  =  12,000  and  a  right  triangle  of  area 
*/2  (600)  X  4  =  1 200,  the  sum  being  13,200. 

Let  y  be  the  horizontal  distance  from  the  vertical  through 
the  center  of  gravity  of  this  trapezoid  to  A.  Taking  moments 
about  A, 

13,200  y  =  12,000  X  2  +  1200  X  ^  X  4  =  27,200 
.*.  y  =  2.07,  or  say,  2.1  ft. 

The  work  has  been  given  in  detail,  in  this  example^  to  give 
the  method  to  follow  in  all  subsequent  examples,  where  only 
results  will  be  stated. 

Let  us  investigate  the  strength  of  the  toe  AL,  at  the  vertical 
section  through  A.  Suppose  reinforcing  rods  to  be  placed,  as 
shown  by  the  dotted  lines,  3"  above  the  base;  then  the  depth 

!d  =  IN  (of  the  formulas  of  the  Appendix)  =  45".    The  bend- 
ing moment  at  A  due  to  soil  pressure  is, 

M  =  13,200  X  2.1  =  27,700  ft.-lbs.  =  332,400  in.-lbs. 
The  percentage  of  steel  to  assume  for  such  a  deep  section 
is  very  small — under  0.2%  perhaps;  hence  from  the  diagram, 
App.,  Fig.  9,  for  0  =  36°  52',  the  inclination  of  the  upper  surface 
of  the  toe  to  the  horizontal,  we  find  j  >  0.92.  Takej*  =  0.92 
.".  jd  =  0.92  X  45  =  41.4  and  assuming  fa  =  16,000  lb./in.2  in 
formula  (14)  of  the  Appendix,  /,  A  j  d  =  M 

.'.  16,000  X  414  A  =  332,400; 
whence  A  =  0.5  sq.  in.  for  b  =  12" '. 

As  in  the  case  of  deep  sections,  the  bond  stress  is  often  large 
with  large  bars,  try  %"  D  bars,  6"  center  to  center  (abbreviated 
hereafter  to  6"  c  to  c  or  6"  £),  corresponding  to  A  =  2  (K)2  = 
0.5  sq.  in.,  for  b  =  12".  The  perimeter  of  one  bar  =  2  =  o\ 
.'.  for  b  =  12",  the  length  of  wall  taken,  20i  =  4. 
By  eq  (34),  App.,  the  bond  stress  is  given  by, 

(j  d  S0i)  «i  =  Q tan  & 

d 

=  13,200  -  332'4°°  X  0.75  =  7660, 
45 

whence,  u\—  =  46.3  lb./in.2 

0.92  X  45  X  4 


132  DESIGN   OF   RETAINING  WALLS 

v  b  Ui'Soi 

Also,  since     HI  =  —  .  .  v  =  

20  b 

whence  the  horizontal  or  vertical  shear  between  the  steel  and 
the  neutral  axis,  is, 

46.3  X  4 
v  =  -  -  =  15.4  lb./m.2 

12 

The  values  of  u\  and  v  are  well  below  the  limits  80  and  40 
lb/in2,  so  that  larger  bars  can  be  used  if  desired.  The  ]/?!'  D 
bars  should  extend  from  near  L,  over  50  diameters  to  the  right 
of  the  vertical  through  A,  or  say  2!  b"  to  right  of  A  to  give 
sufficient  embedment. 

The  value  of  fc  as  computed  from  App.  eq.  15,  is  195  lb./in.2 
For  such  deep  sections,  there  is  no  need  to  compute  fc,  as  it  will 
always  be  less  than  the  allowable  stress,  650  lb./in.2 

84.  Resistance  to  Sliding.  This  subject,  which  should 
generally  be  treated  earlier,  has  purposely  been  deferred,  in 
order  to  make  a  number  of  comments,  which  will  apply  in  other 
designs. 

Calling  /,  the  coefficient  of  friction  of  masonry  on  earth, 
the  product  of  the  total  weight  resting  on  the  foundation  by/ 
must  equal  the  earth  thrust  on  HN  for  exact  equilibrium. 
.'.  39,8oo  /  =  13,730      -.  /  =  0.345.  ^ 

Now  /'  for  moist  clay  is  given  as  0.33 ;  hence  sliding  would 
occur  on  such  a  foundation  or  on  wet  clay  or  quicksand.  This 
may  be  obviated  by  increasing  the  amount  of  masonry,  by 
inclining  the  foundation,  or  often,  by  adding  a  projection  to  the 
footing,  below  it,  and  near  H.  In  the  last  case,  the  wall  cannot 
slide  without  the  projection  pushing  the  earth  before  it;  thus 
substituting  friction  of  earth  on  earth  for  that  of  masonry  on 
earth  which  is  generally  less.  Of  course  the  earth  in  front  of 
the  wall  assists  materially/  when  it  is  well  compacted  after  the 
filling  of  the  trench,  but  it  is  safer  not  to  allow  for  this  in  the 
computation.  , 

For  dry  earth  or  clay,  /  can  be  taken  as  0.5.  Call  <r  the 
factor  of  safety  against  sliding. 

•'•  o-  X  13,73°  =  -o-5  X  39,&>o  =  19,900 
•  '•  o-  =  1.45 


84,  85]      FINDING  LENGTH  OF  T  OR  COUNTERFORTED  WALLS          133 


This  is  perhaps  satisfactory.  In  fact,  if  one  could  be  assured 
that  during  heavy  rains,/7  should  never  fall  below  0.5,  the  depth 
of  footing  could  be  decreased.  However,  it  is  wise  not  to  pare 
down  a  wall  to  a  very  minimum,  since  w,  <p  and/',  vary  according 
as  the  earth  is  dry  or  saturated  and  our  assumptions  too,  as  to 
their  values  and  consequently  the  value  of  the  thrust  may  be 
far  from  the  truth. 

85.  Approximate  Method  of  finding  Length  of  Base  of  T  or 
Counterfeited  Walls.    Let,  Fig.  47,  ABCD  represent  the  base 
slab  and  FI  the  vertical  stem  of  either 
a  T  or  a  counterforted  wall.     IN  is  the 
level  earth  surface  on  which  there  may 
bfe  a  surcharge  of  height  NNf,  which  may 
be  supposed  to  reach  to  either  N  or  /. 


Let  E  =  earth  thrust  on  vertical  plane 
which  acts  a 


D 


E 


•c  \ 


distance  c  above  B.  A 

b  =  length  of  base  AB.  j 

In  computing  the  weight  resting  on 
the  base  AB,  ignore  the  weight  of  toe  FIG.  47 

;  AF  and  for  the  case  where  the  surcharge 
only  extends  to  N,  replace  the  weight  of  the  remainder  of  the 
wall  and  the  earth  over  it  by  that  of  a  rectangular  prism  of 
earth  of  height  BN,  width  GN  and  length  i  foot  perpendicular 
to  the  plane  of  paper.* 

Let  W  =  weight  of  this  rectangular  prism  of  earth  B  G. 
Its  line  of  action  is  thus,  ^  GN  to  left  of  C.  Then  if  R,  the 
resultant  of  E  and  W,  is  assumed  to  cut  AB,  %  b  from  B,  the 
moment  of  W  about  this  point,  will  equal  the  moment  EC  of  E 
about  the  same  point. 

*  This  device  was  suggested  by  an  article  on  the  "  Design  of  Standard  Re- 
inforced-concrete  Retaining  Walls"  in  Engineering  News,  July  24,  1913,  by 
Mr.  H.  M.  Gibb,  who  uses  it  in  deriving  a  number  of  interesting  conclusions. 
The  resultant  on  the  base  (of  length  b)  always  passing  through  the  outer 
third  point,  it  was  found,  as  the  vertical  slab  was  moved  back  from  the  toe 
towards  the  heel,  (i)  that  the  unit  toe  pressures  decrease,  (2)  the  resistance  to 
both  overturning  and  sliding  decreases,  (3)  that  b  is  a  minimum  when  the 
length  of  toe  =  %  b,  but  that  there  is  only  3%  difference  as  the  length  of  toe 
varies  from  >  b  to  b. 


134 


DESIGN   OF  RETAINING  WALLS 


Before  stating  this  equality,  the  desired  length  of  toe  DF, 
must  be  expressed  in  terms  of  b.  The  moment  equation  will 
then  determine  b. 

Ex.  i.  Make  an  approximate  computation  of  the  length  of  base  b  of 
the  T  wall,  Fig.  48,  Art.  86,  the  surcharge  not  extending  to  left  of  N.  From 
the  figure  and  the  computation  for  E  and  c  in  Art.  86,  we  have  for  an  as- 
sumed length  of  toe  DF  =  0.36  b,  GN  =  0.64  b,  Fig.  47;  .'.  W  =  weight 
of  earth  BG  (the  earth  weighing  100  lb./ft.3)  =  0.64  b  X  BN  X  100  =  646 
X  17  =  1088  6-lbs.,  and  it  acts  ^(0.64  &)  =  0.32  b  to  left  of  B  and  0.667  b  — 
0.32  b  =  0.347  ^  to  right  of  point  where  R  cuts  AB,  whence  equating 
moments  of  W  and  E  about  this  point, 

1088  b  X  0.347  b  =  8020  X  7-04  =  56,460 
.*.  bz  =  150  .*.  b  =  12.25',  DF  =  0.36  b  =  4.4' 

It  will  be  observed  that  in  Fig.  48,  b  was  assumed  12.5  ft.,  length  of  toe,  4.5 
ft.  and  in  Fig,  48(0)  (case  a  or  surcharge  extending  only  to  N)  the  resultant 


8'Sufcharge 


>80201bg 


l_tjC 


cuts  the  base  slightly  within  the  outer  third  limit.     The  approximation  is 
thus  remarkably  close  and  it  is  equally  so  in  the  next  example. 

Ex.  2.  Refer  to  Fig.  50,  Art.  86(a)  for  quantities.  Assume  length  of 
toe  =  0.3  b  .'.  GN  =  0.7  b.  Hence  for  surcharge  extending  to  wall,  the 
weight  W  will  be  assumed  to  be  the  weight  of  a  prism  of  earth  BG,  Fig.  47. 
plus  the  weight  of  surcharge,  8'  high,  over  GN  or  the  weight  BN'  X  GN  X  100 


T    WALL    WITH    SURCHARGE 


135 


=  25  X  0.7  b  X  ioo  =  1750  b  pounds.  Here,  W  acts  to  left  of  B,  y£  (0.7  6)  = 
).35  b  and  to  right  of  the  point  %  b  from  B  or  where  R  is  assumed  to  cut 
:he  base  (0.667  —  0.35)  b  =  0.317  b.  Whence,  taking  moments  about  this 
Doint 

1750  X  0.317  b*  =  E  c  =  8020  X  7-04  =  56,460 

.'.  b  =  10.1',  DF  =  0.3  b  =  3.03  ft. 

vhich  agree  almost  exactly,  with  the  values  adopted,  which  correspond  to  a 
•esultant  on  the  base  cutting  it  at  the  outer  third  point.  With  the  surcharge 
extending  only  to  N,  it  will  be  found  that  b  =  12.25'  in  place  of  10.1'. 

!Ex.  3.  Consider  the  counterforted  wall,  Fig.  52,  Art.  88,  where  E  for 
:  ft.  length  of  wall  =  14,660  Ibs.  and  c  =  9.96  ft.  With  the  lettering  of 
"ig.  47,  assume  DF  =  0.3  b  .'.  GN  =  0.7  b  and  for  the  surcharge  extend- 
ng  only  to  N,  W  =  0.7  b  X  BN  X  ioo  =  70  X  25  b  =  1750  b  Ibs.  and 
ts  line  of  action  is  0.35  b  from  B  or  (0.667  —  0.350)  b  =  0.317  b  from  the 
joint  where  R  cuts  the  base,  %  b  from  B. 
Equating  moments  about  this  point, 

I75O  b  X  0.317  b  =  EC  =  146,000 
.'.  bz  =  146,000  -5-  555  =  263  .'.  b  =  16.2  ft., 
tnd  DF  =  0.3  b  =  4.86  ft. 

The  dimensions  actually  adopted,  satisfying  the  outer  third  condition, 
ire  b  =  15.5  ft.,  DF  =  4.5  ft.  =^ length  of  toe. 

It  is  seen  from  these  examples  that  the  approximate  method  gives  rather 
:lose  results.  The  reason  is  (see  Fig.  47)  that  although  W  is  always  less 
han  the  true  weight,  its  arm  about  the  outer  third  point  is  greater  than  the 
rue  arm,  so  that  the  product  is  not  greatly  altered. 


, 


86.  T  Wall  with  Surcharge.     The  length  of  base  shown  in 
7ig.  48,  was  based  upon  the  preliminary  investigation  of  Art.  85, 

Zx.  i. 

MASONRY 

(w'  =  150  Ib.  cu.  ft.) 
Take  moments  about  C 


Section 

Area 

,  Square 

Feet 

Arm, 

Feet 

Area 
Moment 

\bove  CD 

14 
7 
2-33 
2-33 
0.67 

2-33 

X 
X 
X 
X 
X 
X 

I 

I 
2 

3 

12.5 
2.25 

=     7 
=     4 

=    7 

=     8 

=     5 

67 

33 
25 

O 
I 

4 
i 
—  i 

5 
33 

75 
•5 

7- 
9- 
4- 
28. 
14. 
-  7- 

33 
67 

58 
87 

Below  CD 

.  46 

25 

55-71 

Thus  the  weight  of  masonry,  per  foot  of  length,  is  46.25  X  150 
6937  Ibs.  and  its  moment  about  C  is,  55.71  X  150  =  8355 


ft.-lb. 


I 


136 


DESIGN   OF   RETAINING   WALLS 

EARTH  FILLING 

(w  =  100  Ib.  cu.  ft.) 

Take  moments  about  C 


Section 

Area,  Square  Feet 

Arm,  Feet 

Area 
Moment 

Above  D  

7X1=     7- 

1.67 

II  .67 

Below  D  

14  X  6  =  84. 
2-33  X  3  =     7- 

42O. 
42. 

98. 

473-67 

Hence  the  weight  of  earth  over  the  wall,  per  foot  of  its  length 
is  9800  Ibs.  and  its  moment  about  C  is,  47,367  ft.-lbs. 
Case  (a).    With  no  surcharge  over  IN: 


Weight, 
Pounds 

Masonry 6,937 

Earth 9,800 


16,737 


Moment, 
Ft.-Lbs. 

8,355 
47,367 

55,722 


Hence  the  combined  weight  is  16,737  Ibs.  and  it  acts  55,7227 
i6,737  =  3-34  ft.  to  right  of  C. 

Case  (V).  Surcharge  over  IN.  The  weight  of  the  surcharge 
is  7  X  8  X  100  =  5600  Ibs.  and  its  moment  about  C  is  5600  X 
4.5  =  25,200  ft.-lbs.  Add  these  to  the  preceding  totals.  The 
total  weight  of  masonry,  earth  and  surcharge  is  22,337  Ibs.  and 
its  total  moment  about  C  is  80,922  ft.-lbs.  The  resultant  weight 
thus  acts  80,922/22,337  =  3.63  ft.  to  right  of  C.  Railway 
tracks,  13'  c  to  c,  will  be  assumed,  parallel  to  the  wall.  As 
such  tracks  may  be  very  near  the  wall  and  one  or  more  may  be 
occupied  by  locomotives,  both  cases  (a)  and  (b)  of  loading  and 
the  resulting  stresses  should  be  investigated. 

Resistance  to  sliding.     Let/'  =  0.5. 

Case  (a),  factor  safety  =  -  -  =  1.04 


Case  (b),  factor  safety  = 


8020 

0.5  X  22,337 

8020 


]  T    WALL    WITH    SURCHARGE  137 

Face  Slab  CI.  The  height  of  face  slab  is  14  ft.  Let  h  = 
depth  from  /  (top  of  wall)  to  point  considered.  Taking  h  —  15 
gives  a  thrust  in  excess, 

E  =  14.3  (232  -  82)  =  6650  Ibs. 
and  its  resultant  acts  above  the  level  h  =  15, 

/          8x15 

c  =  [i  -\ ) —  =  6.30  ft. 

V        31  '  3 

At  h  =  10,  E  =  14.3  (i82  -  82)  =  3720  Ibs. 
It  acts  above  the  level  h  =  10, 

/          8xio 
V         26'  3 
Similarly  at  h  =  5,  E  =  14.3  (13 2  —  82)  =  1500  Ibs.  and  it  acts, 

/         8x5 

c  —  ( i  H ) —  =  2.30  ft.  above  h  =  5. 

V         2i/  3 

Hence  the  bending  moments  in  the  face  slab, 

at  h  =  15,  =  6650  X  6.3    X  12  =  502,740  in.  Ibs. 

at  h  =  10,  =  3720  X  4.36  X  12  =  194,600  in.  Ibs. 

at  h  =    5,  =  1500  X  2.3    X  12  =    41,400  in.  Ibs, 

These  moments,  like  the  thrusts,  correspond  to  i  ft.  length 

of  wall  .'.  b  =  12". 

Consider  horizontal  sections  of  the  face  slab  at  these  depths, 
for  which  b  =  12"  and  d  =  thickness  of  slab  from  the  front  face 
to  the  center  of  the  steel  bars  placed  parallel  to  and  2.5"  from  ID. 
The  values  of  d  at  h  =  5,  10,  15  ft.  are  13.5,  17.5,  21.5  inches, 
respectively.  Since  the  rear  face  makes  only  a  small  angle  with 
the  exposed  face,  the  formulas  and  diagrams  for  prismatic  beams 
will  be  used,  Appendix,  eqs.  (18),  (19),  (20),  (21)  and  Fig.  n. 

Thus  at  h  =  15,  R  =  —  =       5°2?74°      =  90.5 
bd2       12  X  (21.5)2 

_  M  _         194,600 


ia  X 


138  DESIGN   OF  RETAINING   WALLS 

From  the  diagram,  Fig.  n,  with  R  =  90,  fs  =  16,000,  we 
fmd/c  =  575,  p  =  .0063,  j  =  0.88  .'.  steel  area  at  D  =  p  b  d  = 
0.0063  X  12  X  21.5  =  1.62  sq.  in.  Take  ^"  Q  bars,  4"  c  to 
Cj  giving  A  =  1.68  sq.  in.  for  b  =  12". 

To  test  the  shearing  and  bond  resistance  at  D,  notice  that 
in  formula  (35)  of  the  Appendix,  the  shear  Q  =  earth  thrust. 

Q  6650 

.  .  at  h  =  15,  v  =  —  =  --  =  29  Ib.  sq.  in.; 
bjd       12  X  0.88  X  21.5 

Q  6650 

u  =  -  -  =  -  -  -  =  30  Ib.  sq.  in. 
jdZo       0.88  X  21.5  X3  X3 

Both  are  well  within  limits  prescribed.  We  proceed  similarly 
at  h  =  10. 

With  R  =  53  and  fs  =  16,000,  we  find  from  the  diagram, 
fc  =  420,  p  =  0.0035. 

Since  p  is  about  half  the  previous  value,  cut  out  one-half 

the  bars,  .*.  use  J<"  D  bars,  8"  c  to  c.     This  gives  p  =  —  -f- 

16 

(8  X  17.5)  =  .0040.     With   this  value  of  p  and  R  =  53,  the 
diagram  gives  /,  =  14,600,  fc  =  400,  j  =  0.90. 


For  shear,  v  =     -  =  -        -  =  19.7  Ib.  sq.  in. 

bjd       12  X  .90  X  17.5 


For  bond  stress,  u  =  —  —  =  -   —  -  =  «?2.«;  Ib.  sq.in. 

.go  X  17-5  X  4-5 


Both  are  well  within  the  limits  allowed.      Lastly  at  h  =  5, 

place  y^'   D   bars,   16"  c  to  c.  /.   p  =  —  -5-  (16  X  13.5)  = 

16 

.0026,  and  with  R  =  19,  we  find  from  the  diagrcm, 
/,  =  8000,  fc  =  200,  j  =  0.93 

1500 

.  .  v  =  -  =  10  Ib.  sq.  in. 
12  X  .93  X  13.5 

1500 

u  =  -  -  -  =  53  Ib.  sq.  in. 
.93  X  13-5  X  2.25 

Both  safe  values. 


]  T    WALL    WITH    SURCHARGE  139 

These  (nearly)  vertical  bars  will  be  hooked  in  the  projection 
;  the  bottom  of  the  footing. 

Soil  Pressures.  Recalling  that  in  case  (a),  the  resultant  of 
te  weight  of  wall  and  earth  over  it,  Wa  =  16,737  Iks.,  acts 
34  ft.  to  right  of  C,  and  in  case  (b),  the  resultant,  Wb  =  22,337 
•s.  acts  3.63  ft.  to  right  of  C,  we  proceed  next  to  combine  Wa 
id  Wb  in  turn,  with  the  earth  thrust  E  =  8020  Ibs.  on  BN, 
hich  acts  7.04  ft.  above  B 

Case  (a).  Let  the  resultant  Ra  of  E  and  Wa,  cut  AB  at 
a  (not  shown  on  the  figure)  and  call  the  arm  of  Wa  about 
a  =  da.  Take  moments  about  Fa. 

Wa  da  =  E  X  7.04 
56,460 

Thus  Ra  cuts  the  base  at  Fa,  3.38  -  3.34  =  0.04  ft.  to  left 
F  C,  or  a  =  1.79  ft.  from  the  center  of  the  base  .'.  in  the 
>rmula  for  soil  pressures,  /  =  12.5',  a  =  1.79', 

16,737  /    ±  £^\  =  I34Q  (l  ±  a86) 
12.5    V          12.5  / 

.*.  soil  pressure  at  A  =  2490  Ib.  sq.  ft. 
soil  pressure  at  B  =     190  Ib.  sq.  ft. 

Case  (b).  Let  the  resultant  Rb  of  E  and  Wb,  cut  the  base 
t  Fb,  the  arm  of  Wb  about  Fb  being  db.  Take  moments  about 
\  and  solve  for  db. 


22,337 

Thus  Rb  cuts  AB,  3.63  —  2.53  =  i.io  ft.  to  right  of  C,  or 
..5  +  i.i  =  5.6  ft.  to  right  of  A  or  6.25  —  5.60  =  0.65  ft.  to 
he  left  of  the  mid-point  of  AB. 

.'.  in  the  formula,  a  =  0.65,  /  =  12.5 


12.5   ^         12 


.*.  soil  pressure  at  A  =  2345  Ib.  sq.  ft. 
soil  pressure  at  B  —  1230  Ib.  sq.  ft. 


140 


DESIGN   OF   RETAINING   WALLS 


These  soil  pressures  (or  reactions)  are  plotted  in  Fig.  480 
and  the  resultant  reactions  on  heel  and  toe,  in  position  and 
magnitude,  are  found  as  hitherto  explained  and  marked  on  the 
figure.  The  weight  of  the  heel  (6  ft.  base)  and  earth  over  it, 


Case  (a),  10 730  Iba. 
Case (b), 15 550 Iba. 


h*-  2.33 '-»{  u*.  -2.82  -  ->j 


1230 


FIG.  48(0) 

case  (a)  is  10,750  Ibs.,  for  case  (b),  15,550  Ibs.  The  resultant 
of  each  weight  will  be  regarded  as  acting  along  a  vertical  through 
the  mid-point  of  D'B. 

Heel.  The  heel  DD'B  will  be  treated  as  a  cantilever  beam, 
subjected  to  the  action  of  its  weight,  with  that  of  the  earth  over 
it  and  the  soil  reaction  on  D'B,  the  breadth  of  beam,  perpen- 
dicular to  plane  of  paper,  being  b  =  12  inches. 

Take  moments  about  Df 

Case  (a),  Ma  =  10,750  X  3  —  444°  X  2.26  =  22,216  ft.  Ibs. 
Case  (b),  Mb  =  15,550  X  3  —  9000  X  2.82  =  21,270  ft.  Ibs. 
Ma  is  the  greater  and  will  be  used  in  designing  the  steel-rein- 
forcement, which  is  to  be  placed  3"  from  the  upper  face  of  the 
heel. 


6]  T    WALL    WITH    SURCHARGE  141 

At  the  section  DD',  d  =  distance  from  D'  to  steel,  measures 
,3",  also  0i  =  21°  15',  cos  0i=  0.932,  0!  being  the  angle  the 
od  makes  with  the  normal  to  the  section  DD'  or  with  the 
lorizontal. 

By  App.,  eq.   (n),  M  =fsAi.jd.cos  ft;   whence  assuming 
=  0.9,  fs  =  16,000, 

22,216  X  12  =  16,000  A  X  0.9  X  33  X  0.932 

.".  A  =  0.604  sq.  in. 
If  we  take  $/%'   D   deformed  bars,  6"  c  to  c,  A  =  0.784, 

0.784 

vhence,  p  = =  0.002. 

33  X  12 


From  Fig.  9,  App.,  dotted  lines,  we  find  for  p  =  0.002  and 

=  21°  15',  k  =  0.211,7  =  0.93,  so  that  using  j  =  0.9  above, 
on  the  side  of  safety. 

The  bars  are  to  be  extended  about  40  diameters  =  40  X  5/8 
=  25"  into  the  face  slab  and  hooked  over  a  horizontal  bar  at 
the  end.     This  provides  sufficient  bond  resistance  in  the  vertical 
kkb. 

As  it  is  not  always  possible  to  know  beforehand  the  character 
•of  the  foundation,  it  is  the  practice  of  some  to  omit  the  soil  re- 
action in  the  above  computation.  With  a  yielding  foundation, 
if  the  wall  has  moved  over  at  the  top  so  much  that  there  is  no 
soil  reaction  over  D'B,  not  only  the  weight  of  heel  and  earth 
•over  it,  acts  on  the  heel,  but  in  addition  there  is  the  friction 
due  to  the  earth  thrust  on  BN.  The  latter  is  nearly  equal  to 
Ef  (/ =•  2/3  in  this  case)  and  acts  downward  along  NB.  If 
the  foundation  is  doubtful,  all  of  these  forces  should  be  included 
;in  designing  the  reinforcement.  In  the  Appendix,  Art.  10,  is 
given  an  illustration  of  this  method  of  computation. 

There  is  sometimes  a  difficulty  in  securing  sufficient  bond 
resistance,  due  to  the  short  extension  available  of  the  inclined 
bar  into  the  face  slab.  This  may  be  met  by  bending  the  bar 
downward  across  the  stem  and  into  the  toe. 

Maximum  unit  shear  and  bond  stress  at  DD'.  The  external 
shear  is  greatest  for  case  (b)  and  is, 

Q  =  1 5,550  -  9000  =  6550  Ibs. 


142  DESIGN   OF  RETAINING   WALLS 

since  ft  =  o,  formulas  (20)  and  (33),  App.,  reduce  to, 

M 

HI  .jd2oi  =  Q tan  fti  =  vb  .jd. 

d 

Ignoring  the  term  M  tan  fti/d,  the  resulting  values  of  u\  and 
i)  will  be  in  excess. 


Shear  *=-£-=  ^ =  18  lb./in.2 

0.yd       12  X  0.93  X  33 

With  f£"  n  bars,  6"  c  to  c,  0i  =  2.5  in  6  in.  .'.  20i  =  5  in 
12  inches.  Hence  the  bond  stress 

=  42.6  lb./in.2 

jd  20i      0.93  X  33  X  5 

Hence,  both  shear  and  bond  stresses  are  safe.  The  unit 
stress  fc  in  the  concrete  at  D' ',  was  computed  and  found  to  be 
small. 

Toe  AC'C.  We  have  cos  ft  =  cos  21°  15'  =  0.932,  cosz  ft  = 
0.869.  Assume  k  =  0.23  .'.  j  =  0.92,  corresponding  to  p  = 
0.002  by  Fig.  9,  App.,  .'.  kj  =  0.212. 

As  in  the  previous  design,  neglect  the  weight  of  toe  for  addi- 
tional security  and  compute  shear  and  moment  at  CCf  from  the 
soil  reaction  only.  Both  are  greatest  for  case  (0),  for  which 
Q  =  9590  Ibs.  and  M  =  9590  X  2.33  X  12  =  268,140  in.  Ibs. 

Assume  horizontal  bars,  3"  from  AB  and  extending  7  ft. 
from  A  .'.  at  CC',  d  —  33".  Assuming  fs  =  16,000  lb./in.2,  by 
App.,  eq.  (14),  steel  area 

268,140 

A  =  — —  =  0.554  sq.  in. 

16,000  X  0.92  X  33 

Here,  either  J^"  D  bars,  12"  c  to  c,  or  5/8"  D  bars,  6"  c  to  c, 
will  suffice.  Let  us  test  the  latter  for  shear  and  bond  stress. 
For  the  $/%'  Q  bars,  6"  c  to  c,  20X  =  2  X  2.5  =  5  sq.  in.  By 
App.,  eq.  (34), 

M 

HI  .jd  20i  =  Q tan  ft  =  vb  .jd. 


86]  T    WALL    WITH    SURCHARGE  143 

Neglecting  the  term  in  M  to  give  an  excess. 

.*.  bond  stress  u\  —  -  =  63  lb./in.2 
0.92  X  33  X  5 


unit  shear  v  =  -        -  =  26,5  lb./in.2 
12  X  0.92  X  33 

As  these  are  safe  values,  the  true  shear  and  bond  stresses, 
which  are  less,  are  safe  values. 

For  the  yi"  D  bars,  6"  c  to  c,  A  =  0.784  sq.  in.  .*.  p  = 
0.784  -T-  (12  X  33)  =  0.002,  as  assumed  above.  Also  from 
App.  eq.  (15),  it  is  found  that  fe  =  224  lb./in.2 

Let  us  now  test  the  yj'  D  bars,  12"  c  to  c,  for  bond  and 
shear  stresses.  Here  the  term,  M  tan  &/d  must  be  included. 

M  268,140 

We  have,  —  tan  /3  =  -  X  0.389  =  3160  Ibs. 
*  33 

ui.jd  S0i  =  9590  —  3160  =  vb.jd. 
Since  S0i  =  3. 

6430 


0.92  X  33  X  3 
6430 


=  70.7  lb./in2, 
=  18  lb./in.2, 


12  X  0.92  X  33 

,. 

both  safe  values,  so  that  $/%'  D  bars,  12"  c  to  c,  can  be  used 
if  desired. 

Temperature  Reinforcement.  The  area  of  ICD,  Fig.  48,  is 
18  X  168  =  3024  sq.  in.  and  y$  of  i%  of  3024  =  10.08  sq.  in. 
22  —  yj'  D  bars,  give  a  total  area  =12.4  sq.  in.  Hence  for 
temperature  reinforcement,  use  14  horizontal  yjr  D  bars,  12" 
c  to  c  and  2%"  from  the  exposed  face  and  8  horizontal  yj'  D 
bars,  24"  c  to  c,  resting  against  the  vertical  reinforcement  of 
the  vertical  slab  or  3"  about,  from  rear  face. 

The  horizontal  bars  near  the  front  face  rest  against  y^"  D 
vertical  bars,  30"  c  to  c.  Also  yi"  D  horizontal  bars,  spaced 
12"  c  to  c,  will  be  placed  longitudinally  or  parallel  with  the  face 
wall  under  the  inclined  rods  of  the  heel  and  over  the  horizontal 
rods  of  the  toe. 


144 


DESIGN   OF   RETAINING  WALLS 


In  the  final  drawing,  Fig.  49,  all  the  rods  are  shown  in  posi- 
tion, and  it  will  be  noticed  that  a  projection  on  the  bottom  of 
the  footing  has  been  added,  not  only  to  furnish  additional  bond 
resistance  to  the  vertical  reinforcement,  but  also  to  increase 
materially  the  resistance  to  sliding.  Before  sliding  can  occur, 
the  projection  will  have  to  push  the  earth  to  the  left.  Hence 
for  the  5.5  ft.  from  A  to  the  projection,  on  which  the  soil  re- 


T"T 


FIG.  49 


action,  case  (a),  is  easily  found  from  Fig.  49,  to  be  10,890 
the  coefficient  of  friction  is  that  of  earth  on  earth.     Hence  the 
resistance    to    sliding   is    ^  X  10,890  =  7260    Ibs.      The    soi 
reaction  on  the  remainder  of  the  base  is  5847  Ibs.,  and  the  tofc 
frictional   resistance,    for  /  =  0.5,   is   0.5  X  5847  =  2923    11 
Thus  the  total  resistance  to  sliding  is  7260  +  2923  =  10,183. 


J,  86  (a)] 


T    WALL    WITH    SURCHARGE 


145 


>ince  the  horizontal  earth   thrust  =  8020  Ibs.,   the  factor  of 


fety  against  sliding  = 


10,183 
8,020 


1.27, 


Thus  the  projection  has  increased  the  resistance  to  sliding 
>y  22%.    To  be  most  effective,  it  should  be  placed  as  far  to  the 


FIG.  50 

!rear  as  possible  and  not  under  the  toe.  The  earth  in  front  of 
the  wall  offers  some  resistance,  especially  if  it  is  well  rammed, 
but  it  is  well  to  ignore  it  in  the  computation. 

It  will  be  observed  that  the  projection  is  under  "punching" 
shear,  where  120  Ib.  sq.  in.  is  allowed.  The  resistance  at  a 
horizontal  section  is  thus  24  X  12  X  120  =  29,760  Ibs.,  which 
is  much  greater  than  the  shearing  force  acting  on  it. 

86  (a).  A  second  familiar  type  of  T  wall  is  shown  in  Fig.  50, 

*  This  is  doubtless  too  high;  porous  earth  yields  more  readily  than  a  solid 
to  compression,  and  any  sliding  would  be  progressive,  starting  first  at  the 
projection. 


146  DESIGN   OF   RETAINING   WALLS 

the  total  height  of  wall  and  surcharge  being  the  same  as  for  the 
type  Fig.  48.  The  surcharge  for  this  wall  is  supposed  to  cor- 
respond to  a  fixed  load,  as  buildings,  machinery,  etc.,  and  its 
weight  is  supposed  to  correspond  to  that  of  earth,  8  ft.  high, 
extending  up  to  the  wall  and  fixed  in  position.  The  front  face 
of  the  stem  of  the  T  is  vertical,  the  thickness  at  top  is  i  ft.  and 
the  rear  face  produced  meets  AB  at  J,  5  ft.  to  the  right  of  A 
and  5  ft.  to  left  of  B.  The  preliminary  computation  for  the 
length  of  base  has  been  given  in  Art.  85,  Ex.  2.  The  student 
can  show  that  with  the  dimensions  given  in  Fig.  50,  the  resultant 
on  the  base  cuts  it  almost  exactly  y$  L  H  from  Z,,  giving  a  vertical 
component  of  soil  pressure  at  L,  3970  lb./ft.2  and  at  H,  zero. 
The  factor  of  safety  against  sliding,  for/'  =  0.5,  is  1.24,  not  in- 
cluding the  aid  given  by  the  projection. 

The  reinforcement  throughout  was  placed  2.5  inches  from 
the  nearest  face.  That  for  the  vertical  slab  is  precisely  the 
same  as  for  the  preceding  design,  the  stem  being  practically 
the  same.  The  reinforcement  for  the  fillets  and  base  slab  are 
treated  in  full  in  the  Appendix,  Arts,  n  and  18. 

If  preferred,  the  upper  reinforcement  of  the  rear  fillet  can 
be  bent  downward  across  the  stem  and  be  made  to  pass,  say 
2^2",  from  the  upper  surface  of  the  front  fillet,  thus  materially 
ensuring  it  against  shrinkage  cracks.  Similarly,  it  is  sometimes 
the  practice  to  bend  some  of  the  vertical  bars  of  the  rear  face 
of  the  vertical  slab  across  the  base  slab,  the  continuation  extend- 
ing, say  2.5  ins.,  from  LH  and  constituting  the  lower  reinforce- 
ment of  the  base  slab. 

87.  T  Wall.  Surface  of  Earth  Sloping  Upward.  When  the 
surface  of  the  earth  slopes  indefinitely  upward  at  any  angle  less 
than  or  equal  to  <p,  the  earth  thrust  on  .BAY  Fig.  51?  w^  ac^ 
parallel  to  the  top  slope.  Next  the  face  slab,  since  the  earth 
generally  settles  relatively  to  the  wall,  friction  of  earth  on  wall 
will  be  exerted,  and  the  earth  thrust  will  be  changed  in  amount 
and  direction,  except  for  the  case  where  the  surface  slopes  at 
the  angle  of  repose.  For  simplicity,  suppose,  for  any  inclination 
of  the  surface,  that  the  earth  thrust,  for  the  same  depth,  remains 
the  same  in  amount  and  direction  as  the  wall  is  approached. 


S7j 


T  WALL.       SURFACE    OP  EARTH    SLOPING   UPWARD          147 


In  Fig.  51,  draw  D'M  parallel  to  IN  and  let  E'  =  earth  thrust 
on  MN  =  thrust  on  D'l,  E"  =  earth  thrust  on  EM,  both 
acting  parallel  to  the  surface  IN. 

.'.  E  =  earth  thrust  on  BN  =  E'  +  E". 
Also,  let  W  =  weight  of  wall. 

W"  =  weight  of  earth  over  heel. 

When  E'  acting  on  ID'  and  E"  acting  on  EM  are  combined 
with  W',  the  resultant  on  the  base  is  the  same  as  the  resultant 
of  E  =  E'  +  E",  acting  on  BN  and  W.     Such  a  resultant,  for 
usual  designs,  will  cut   AB  so  near  A 
that  the  wall  will  tend  to  move  over  at 
the  top.    This  tendency  is  counteracted 
by  W" ',  which  thus  inevitably  comes  into 
play,  so  that  the  true  resultant  on  the  base 
is  found  by  combining  E  with  W'  and 
W".    Thus  the  usual  procedure  is  justi- 
fied.   A  special  case  is  when  IN  is  hori- 
zontal, as  in  previous  examples. 

For  the  T  wall  of  Fig.  51,  the  face  slab 
is  designed  as  before,  using  the  horizontal 
component  of  E'  and  ignoring  the  vertical  component.  The 
soil  reactions  are  found  as  usual,  as  soon  as  the  resultant  of  E, 
Wf  and  W"  is  found  and  the  point  where  it  cuts  the  base. 
Then  the  toe  is  designed  as  hitherto  explained. 

For  the  heel,  note  that  the  inclined  thrust  Ef  on  MN  is  trans- 
mitted unchanged  to  D'l.  Its  vertical  component  is  not  then 
transmitted  to  E  if  the  wall  and  foundation  are  absolutely 
unyielding.  The  thrust  E"  is  carried  directly  to  the  heel  and 
its  moment  about  the  mid-point  of  the  section  DD' ',  must  be 
added  algebraically  to  the  moments  due  to  the  total  soil  reaction 
on  DB,  the  weight  of  the  heel  D'B  and  the  weight  W"  of  the 
earth  over  it.  But  since  the  wall  and  its  foundation  are  not 
rigid,  there  will  be  a  moving  over  of  the  wall  from  the  more  or 
less  elastic  yielding  of  wall  and  foundation,  so  that  a  certain 
unknown  amount  of  friction  will  be  exerted  along  BN,  tending 
to  counteract  the  motion.  The  full  amount  of  friction  that 
can  be  exerted  is  approximately  /  times  the  horizontal  com- 


148  DESIGN    OF   RETAINING   WALLS 

ponent  of  E3  and  all  of  this  will  come  into  play  if  the  wall  is 
on  the  point  of  overturning. 

As  the  foundation  is  often  doubtful,  it  seems  best  to  design 
the  heel  for  this  extreme  case,  and  it  is  suggested  that  the  re- 
inforcement at  DD'  be  figured  from  the  three  vertical  forces: 
the  vertical  component  of  the  soil  reaction  .on  DB,  the  weight 
W"  and  the  full  friction  that  can  be  exerted  along  NB.  This 
neglects  the  friction  on  the  base  and  the  horizontal  component 
of  En ',  which  is  on  the  side  of  safety. 

It  may  be  thought  that  the  horizontal  component  of  E, 
acting  to  the  left,  at  a  distance  y$  BN  above  J2,  should  be 
included  in  the  computation,  but  it  must  be  borne  in  mind  that 
the  earth  over  the  heel  is  not  a  solid  body,  rigidly  connected 
with  the  heel,  but  a  granular  mass;  so  that  the  only  way  to 
transfer  even  the  horizontal  component  of  the  earth  thrust 
acting  on  B'M  is  by  friction,  exerted  along  the  top  of  the  heel, 
between  the  heel  and  the  earth  over  it.  Also,  by  assumption  the 
horizontal  component  of  E',  acting  on  MN,  is  supposed  trans- 
mitted to  D'I\  hence  it  should  be  omitted  from  the  forces  as- 
sumed to  act  on  the  he 

Since  it  cannot  be  known  in  advance  how  much  friction  is 
exerted  on  BN,  it  is  sometimes  omitted,  as  well  as  the  soil 
reaction,  so  that  the  heel  is  designed  simply  foT  the  force  W". 

Railroad  Embankment.  When  the  T  wall  sustains  a  railroad 
embankment,  sloping  up  from  the  top  of  the  wall,  the  surface 
at  a  certain  height  being  level  to  accommodate  tracks,  only  an 
approximate  solution  can  be  indicated.  The  tracks  will  be 
supposed  to  carry  trains,  or  for  computation,  an  equivalent 
weight  of  earth.  Then  the  construction  of  Ar+  32  is  used  to  find 
the  thrust  on  BN  and  MN,  assuming  an  inclination  to  the  thrust 
intermediate  between  IN  and  the  horizontal.  For  safety,  suppose 
any  thrust  to  act  4/IO  height  above  the  base.  See  Art.  43 

The  earth  thrust  on  D'l,  Fig.  51,  will  be  taken  as  the  same 
in  amount  and  direction  as  that  on  MN.  The  design  of  stem, 
toe,  and  heel  then  proceeds  exactly  as  before.  The  inclination 
of  the  thrust  on  BN  or  MN  can  not  be  indicated  exactly.  For 
no  surcharge  the  thrust  is  horizontal;  for  a  high  embankment 


87,88]  COUNTERFORTED    WALLS.       FACE    SLAB  149 

(100  to  150  ft.)  and  low  retaining  wall  (10  to  20  ft.)  the  thrust 
makes  nearly  the  angle  y  with  the  horizontal.  For  intermediate 
heights,  its  inclination  lies  between  o  and  <p.  A  safe  value  can 
be  chosen.  For  very  high  embankments,  the  solution  approaches 
that  for  an  indefinitely  great  surcharge. 

88.  Counterforted  Walls.  For  retaining  walls  of  heights 
greater  than  about  17  ft.,  counterforted  walls  are  found  to  be 
more  economical  than  T  walls. 

In  Fig.  52  is  shown  this  type  of  wall  in  plan  and  elevation. 
In  the  elevation,  FI  is  the  face  slab,  AF  the  toe,  DB  the  heel 
slab,  with  the  "projection"  extending  2  ft.  below  B.  The 
counterforts  DC  I  are  18  ins.  wide  and  10  ft.  apart  center  to 
center.  A  coping  should  be  added  at  the  top  of  the  exposed 
face,  in  this  and  previous  walls,  but  it  is  purposely  omitted  to 
simplify  computations.  The  dimensions,  shown  fully  on  the 
figure,  were  only  arrived  at  after  a  preliminary  design  with  a 
shorter  base  was  investigated  and  found  insufficient,  giving  too 
low  a  resistance  to  sliding  and  too  large  soil  pressure  at  the  toe. 
Also  see  Art.  85,  Ex.  3.  The  thickness  of  face  slab  was  ar- 
bitrarily taken  at  12  ins.,  since  thinner  walls  are  not  so  easy 
to  pour. 

A  surcharge  of  loads  equivalent  to  8'  depth  of  earth  is  as- 
sumed. The  wall  will  be  designed  for  reinforcement  for  the 
same  working  stresses  and  data  used,  as  in  previous  computations : 
fs  =  16,000  Ibs.  sq.  in.,  fc  =  650  Ibs.  sq.  in.,  v  =  40  Ibs.  sq.  in., 
u  =  80  Ibs.  sq.  in.  for  plain  bars,  n  =  15,  <p  =  33°  41',  w  =•  100 
Ibs.  cu.  ft.,  w'  =  150  Ibs.  cu.  ft.,  soil  pressure  =  5000  Ibs.  sq.  ft., 
and  the  coefficient  of  friction/'  of  masonry  on  earth  =  0.5. 

Face  Slab.  The  face  slab,  of  course,  acts  as  a  whole,  but  for 
purposes  of  computation  it  will  be  supposed  divided  into  a 
series  of  independent  continuous  horizontal  beams,  each  12" 
•high,  10'  span  (c  to  c  of  counterforts),  with  bending  moments 

at  center  and  at  junction  with  counterforts  of  —  pi2  and 

12  12 

pi2  (in  ft.  Ibs.),  where  p  —  horizontal  thrust  of  the  earth  in 
Ibs.  per  sq.  ft.,  at  the  depth  h  from  top  of  wall  to  center  of  beam 


i  • 


150 


DESIGN   OF  RETAINING   WALLS 


and  I  —  length  of  beam  =  10  ft.     This  is    decidedly    on    the 
side  of  safety  near  the  junction  of  the  face  slab  with  the  footing, 


Counterfort 

i 

12'-^ 

A 
<=- 

jto          • 

00 

<—«'«-'-*• 

Heel  Slab 

8 

1 

H 

«--      --10^-- 

-  _—  * 

PLAN 


:i: 


FIG.  52 


which  is  not  undesirable.  The  reinforcing  bars  will  be  placed 
2"  from  the  front  face  for  its  whole  length  and  other  bars  will 
be  placed  2"  from  the  rear  face  at  the  counterforts  and  ex- 


COUNTERFORTED    WALLS.       FACE    SLAB 


151 


tending  to  %  span  on  either  side,  to  take  care  of  the  negative 
moments. 

The  surcharge  being  8  ft.,  the  unit  pressure  p  at  any  depth 
//  below  the  top  of  the  wall  is,  Art.  51, 

p  =  w  (8  +  k)  tan2  (45°-  ' -)  =  28.6  (8  +  h). 
The  moment  at  center  and  ends  of  beam  is,  in  amount, 
M  =  __ p^  x  12  =  loop,  in.  Ibs. 

12 

It  will  be  found  that  the  spacing  of  the  horizontal  reinforce- 
ment is  really  determined  by  the  bond  stress,  as  given  further 
on;  hence  the  approximate  formula  for  A  was  used. 

MS  ~  fsjd  A  =  16,000  X  TA  d  A  =  14,000  d  A. 

•••*  =  —, 

14,000  d 
where  d  =  12  —  2  =  ioin.  throughout. 

The  size  and  spacing  of  the  bars,  as  determined  from  this 
formula,  are  given  in  the  following  table: 

HORIZONTAL  BARS  IN  VERTICAL  SLAB 


Depth 

P, 

M  =  100  p 

REINFOF 

.CEMENT 

£ 

Pounds 
Sq.  Ft. 

Inch- 
Pounds 

1             M 

Adopted 

14000^ 

Spacing 

2  

286 

28,600 

o  204. 

1A'   D 

i4/  i 

8"  i 

7  

4.2Q 

4.2  QOO 

o  ^07 

y*   a 

10'   i 

8"  i 

12  

572 

57  2OO 

O   4.OQ 

V2'     D 

f  t 

l"l 

17  

715 

7  I.  5OO 

O    511 

X'    o 

5'   i 

5"  i 

22  

858 

85,800 

o.6n 

y2'  a 

4'   \ 

4"  i 

Spacing  for  Bond  Stress.  Taking  the  working  value  for 
bond  stress,  u  =  So  Ibs.  sq.  in.,  by  App.,  Art.  20,  the  spacing 
in  inches  is,  for  b  ='  12", 

960  jd 


where  Q  =  external  shear. 


152 


DESIGN   OF   RETAINING   WALLS 


For  a  yi"   D   bar,  perimeter  =  (o)  =  2 ;  also  jd 

16,800 


. .  x  — 


Q  =  shear  =  reaction  of  beams  10'  long,  i'  high 

=  y2px  10  =  $p* 


h, 
Ft. 

Q  =  SP 
Lbs. 

16800 

*=-Q- 
In. 

2                                                                           . 

I  4^O 

12"  4 

7 

2.I4C 

8"  ^ 

12  

2,860 

6"  4 

17  .  . 

•l  eye 

s"  i 

22  

4..2QO 

A"  <k 

Some  of  these  values  being  less  than  those  found  above 
were  adopted,  as  indicated  in  the  table  above. 

The  values  of  fc  for  any  h  are  all  less  than  650  Ibs.  sq.  in. 
as  can  be  shown  most  easily  by  aid  of  App.,  Fig.  ii. 

M  M  M 

Thus,  R  =  —  = 


12  X  10' 


1 200 


Hence  at  h  =  22,  R  =  71.5  and  with/5  =  16,000,  the  diagram 
gives,  p  =  0.005,  fc  =  500  Ibs.  sq.  in.,  j  =  0.89  .'.  A  =  p  .bd  = 
0.600  sq.  in.  The  maximum  shear  is  at  h  =  22  and  is, 


v  = 


Q 


4290 


=  40  lb.  sq.  in. 


bjd       12  X  9 

which  is  allowable.     See  p.  164,  for  increase  in  steel  area  due 
to  temperature  stresses. 

Earth  Thrust  on  10  ft.  Length  of  Wall.     The  horizontal  earth 
thrust  on  BN,  for  10  ft.  length  of  wall  is, 

E  =  y2  w  tan2  (45°  -  — )  X  (33'  -  82)  X  10  =  146,580  Ibs. 
Otherwise:  at  N,  p  =  228.8  Ibs.  sq.  ft., at  £,  p  =  943.8  Ibs. sq.ft. 
.'.  E  =  y2  (228.8  +  943.8)  X  25  X  10  =  146,580  Ibs. 

*  Strictly,  the  value  of  Q  should  be  computed  for  beams  only  8'  6"  long. 


COUNTERFORTED    WALLS.       EARTH    THRUST 


153 


This  thrust  acts,  Art.  51, 
h 


8 


)     =  9.96  ft. 


3  25  + 

ibove  B. 

Resultant  of  Weights  in  Magnitude  and  Position.  The  result- 
ants of  the  weights  of  concrete  and  rilling,  with  and  without 
surcharge  over  IN,  must  next  be  ascertained  for  10  ft.  length 
:>f  wall,  including  one  counterfort.  The  filling,  to  the  left  of  the 
Diane  BN,  consists  of  a  triangular  prism  over  the  counterfort 
and  a  rectangular  prism  between  the  counterforts.  The  con- 
crete section  is  divided  up  as  indicated.  Moments  were  taken 
about  D. 

WEIGHTS  AND  MOMENTS  ABOUT  D 


Prism 


Length, 
Feet 


Area  of 
Cross- 
Section, 
Square  Feet 


Weight, 
Pounds 


Arm, 
Feet 


Moment, 
Foot-Pounds 


Filling 

Filling 

Surcharge.  . 
Wall 

DB 

Projection. . 
Counterfort 

Stem 

Toe 

Toe.. 


10. 

IO. 
10. 

i- 
10. 
10. 

IO. 


22. 5X   5 

22.5X10 

8X10 

2.5X10 

2X     2 

22. 5X   5 
25     X    i 

i      X4-5 
2.5X2.25 


16,875 

191,250 

80,000 

37,500 

6,000 

25,312 

37,5oo 

6,750 

8,437 


+6.67 
+5- 

+5- 


+5. 
+9 
+3. 
— o, 

-3 

—  2. 


+  112,500 

+956,250 

+400,000 

+  187,500 
+  54,000 
+  84,373 

-  18,750 

-  21,937 

-  21,094 


With  surcharge  over  IN,      Wz  =  409,624,  Mz  =  1,732,842. 

Without  surcharge  over  IN,  Wi  =  329,624,  Mi  =  1,332,842. 

The  surcharge  here  replaces  the  train  loads,  which  may 
occupy  one  or  several  tracks,  parallel  to  the  wall.  Thus, 
without  ultra  refinement  as  to  the  exact  position  of  the  tracks, 
the  surcharge  will  be  supposed  to  extend  to  BN  or  to  7, 
according  as  the  one  or  the  other  position  will  give  the  largest 
stresses  at  any  section  considered.  Fig.  53. 

The  distance  from  D  to  the  line  of  action  of  the  resultant  of 
the  combined  weights  is: 


Case  (a),  without  surcharge  over  IN  = 


1,332,842 
329,624 


=  4.04  ft., 


154 


DESIGN   OF  RETAINING  WALLS 


Case  (b),  with  surcharge  over  IN       =  — =  4.23  ft. 

409,624 

Thus  the  resultant  in  Case  (a)  acts  9.54  ft.  from  A,  in  Case 
(b)  9.73  ft.  from  A. 

In  Case  (a),  let  RI  denote  the  resultant  of  E  and  W\\ 


1021 


FIG.  53 


In  Case  (6),  let  R2  denote  the  resultant  of  E  and  W 

also  let  0i  denote  the  angle  RI  makes  with  the  vertical; 

let  62  denote  the  angle  R%  makes  with  the  vertical. 


E         146,600 

.'.  tan  61  =  —  =  

Wi      329,624 


E 

tan  62  =  — 


146,600 


0.445; 


=  0.358. 


409,624 

From  Fig.  53  it  is  seen  that  RI  cuts  AB}  9.54  —  9.96  tan  61 
=  9.54  —  4.43  =  5.11  ft.,  to  the  right  of  A  and  Rz  cuts  AB, 
9-73  —  9-96  tan  62  =  9.73  —  3.56  =  6.17  ft.  to  the  right  of  A 
(R%  was  not  drawn  to  avoid  confusing  the  figure). 

Resistance  to  Sliding.    Let  <TI  be  the  factor  of  safety  against 


]  COUNTERFORTED    WALLS.       SOIL   PRESSURES  155 

iding  in  Case  (a);  0-2  the  factor  in  Case  (b)  .'.  0-1  E  =  j'W\ 

E  f          0.500 

.'.  0-1 —  =  o-i  tan  0i  —  f    .'.  ffi  =  =  =  1. 12 

Wi  tan  0i       0.445 

/'          0.500 

imilarly  in  Case  (b),  v2  =  -   —  =  =  1.40. 

tan  02       0.358 

The  increase  in  these  factors,  due  to  the  projection  at  B,  is 
larked  and  can  be  estimated  as  in  the  previous  example. 

Soil  Pressures.  In  Case  (a),  RI  cuts  AB  practically  at  the 
:iird  point;  hence  the  unit  pressure  at  B  is  zero  and  at  A 
ouble  the  mean;  or  since  W\  corresponds  to  10  ft.  length  of 

all,  the  pressure  at  A  =  2  X    329?  24    =  4253  Ibs.  sq.  ft. 

15.5  X  10 

In  Case  (b),  R2  meets  AB,  7.75  -  6.17  =  1.58  ft.  to  the 
:ft  of  the  center  of  AB  .'.  /  =  15.5,  a  =  1.58. 

40,962  /      _   9.48 


( i  =*=  — — )  =  2640  (i  ±  0.612) 

V  TC.C/ 


=  4256  Ibs.  sq.  ft.  at  A,  1024  Ibs.  sq.  ft.  at  B. 

These  pressures,  for  both  Cases  (a)  and  (b),  are  laid  off  as 
rdinates  in  Fig.  53  and  the  diagrams  of  soil  pressures  'completed 

shown.  The  unit  pressures  vertically  under  F,  D,  etc., 
rere  measured  to  scale  and  marked  on  the  figure. 

The  soil  pressure,  at  any  particular  distance  from  A,  is  the 
verage  for  the  10  ft.  length  of  wall.  Under  the  toe,  the  pres- 
are  is  probably  uniform  along  a  line  at  the  same  distance 
ciroughout  from  the  front  edge  of  the  footing,  but  along  such 
line  of  the  heel  slab  the  soil  pressure  is  least  under  the  counter- 
3rts  and  greatest  midway  between  counterforts,  particularly 
Dward  the  rear  of  the  heel  slab.  Reasons  will  be  given  for  this 
lew  when  the  heel  slab  design  is  taken  up. 

Design  of  Toe.  Take  the  section  at  F,  whose  vertical  height 
i  42"  and  breadth  b  =  12"  perpendicular  to  the  plane  of  the 
•aper.  The  total  upward  acting,  soil  reaction  on  the  toe,  for 
;  ft.  length  of  wall  is  greatest  for  Case  (b)  and  =  %  (4256  +  333°) 
<  4.5  =  17,100  Ibs. 


156  DESIGN   OF   RETAINING   WALLS 

Its  resultant  acts  2.3  ft.  to  the  left  of  F;  hence  the  bending 
moment  at  the  section  at  F  is, 

M  =  17,100  X  2.3  X  12  =  471,960  in.-lbs. 

The  shear  at  the  section  at  F  =  17,100  Ibs.  =  Q. 

The  upper  surface  of  the  toe  makes  the  angle  /3  =  29°  of  with 
the  horizontal.  For  an  assumed  steel  percentage  0.2,  from 
App.,  Fig.  9,  we  find  j  =  0.92.  Assume  the  steel  reinforce- 
ment 3"  from  AB\  then  d  =  42  —  3  =  39",  jd  =  35.9.  From 
App.,  eq.  (14),  assuming  fs  =  16,000  lb./in.2, 

M  471,96° 

A  =  -  =  -  -  =  0.822  sq.  in 

fg.jd      16,000X35.9 

$/%'  D  bars,  5"  c  to  c,  giving  A  =  0.9408   for   b  =  12",  will  be 
used.     Then,  p  =  A/bd  =  0.002,  as  assumed. 

M  tan  |8  47I>96°  X  0.557 

Q  ---  =  17,100  --  -  =  10,360  Ibs. 

d  39 

.'.  App.,  eq.  (34), 

0i  =  10,360  =  vb.jd 


For  b  =  12",  201  =  --  (2.5)  =  6. 

.'.  bond  stress  u\  —  --  -  --  =  48.2  lb./in.2, 
35-9  X  6 

shear  v  =  -  —  ^  --  =24  lb./in.2, 
12  X  35-9 

both  safe  values.  In  fact,  since  the  weight  of  toe  and  reaction 
friction  on  base  were  omitted  in  computing  Q  and  M,  the  true 
values  of  A,UI  and  v,  are  all  less  than  the  above. 

The  $/%"  D  bars,  5"  c  to  c,  will  be  placed  3"  above  the 
base  AB  of  the  footing,  every  third  bar  extending  13.5  ft.  from 
A  ,  the  remaining  bars  7  ft.  from  A  . 

Heel  Slab.  The  heel  slab  between  counterforts  up  to  "pro- 
jection" is  supported  on  three  sides  by  the  face  slab  and  counter- 
forts and  to  some  extent  on  the  rear,  by  the  stiffer  beam  over 


COUNTERFORTEDS  WALLS.   HEEL  157 

d  including  the  projection.     Other  considerations,  too,  add  to. 
e  complexity  of  a  solution.     The  thrust  of  the  earth  against 
e  face  slab  is  first  carried  by  beam  action  to  the  counterforts, 
u'ch  rotate  slightly  under  the  pulls.     The  consequent  upward 
ovement  at  the  rear  tends  to  lift  the  entire  heel  slab.     How- 
er,  the  part  of  this  heel  slab  between  counterforts  is  very  much 
ore  .flexible  than  the  remainder  joined  to  the  stiff  counterfort, 
nee  it  will  tend  to  deflect  more.     But  this  increased  deflection 
at  once  opposed  by  an  increased  soil  reaction.     Hence  if  we 
:ippose  a  beam  cut  out  from  the  heel  slab  by  planes  i  ft.  apart 
id  parallel  to  the  face  slab,  as  shown  by  the  dotted  lines  Fig. 
>  (plan),  the  soil  reaction  for  such  a  beam,  10  ft.  long,  center 
center  of  counterforts,  has  an  average  value  given  by  the 
•dinate  under  its  center  of  the  soil  pressure  diagram,  Fig.  53, 
at  the  unit  soil  reaction  is  less  than  the  mean,  under  the  counter- 
•rt  and  greater  than  the  mean  near  the  center  of  the  beam, 
i  fact,  it  is  probable,  for  ordinary  designs,  that  the  soil  reaction 
nder  the  "projection"  at  B  may  be  nothing  at  the  counterforts 
;id  sufficient  at  the  center  to  prevent  any  considerable  deflection 
I  the  supposed  beam.     The  assumption  of  a  uniform  soil  pres- 
iire  over  a  beam,  such  as  is  indicated  in  Fig.  52  (plan)  by  the 
otted  lines,  is  thus  on  the  side  of  safety. 

As  a  basis  for  a  practical  computation,  such  beams  will  be 
•eated  as  independent,  continuous  beams,  with  moments  at 
ic  middle  and  ends  of  /{2  rf,  —  T/I2  wl2,  respectively.  Consider, 
rst,  the  beam  cut  from  the  heel  slab  next  the  projection,  i  ft. 
ide,  /  =  10  ft.  long  and  30  in.  deep. 

From  Fig.  53,  the  soil  reaction  in  Case  (a)  is  700  and  in  Case 
b)  1550  Ibs.  sq.  ft.  In  Case  (a)  the  weight  of  earth  above  the 
earn  is  22.5  X  i  X  100  =  2250  Ibs.  sq.  ft.,  and  in  Case  (b) 
0.5  X  i  X  ioo  =  3050  Ibs.  sq.  ft.  The  weight  of  beam  per 
near  foot  is  2.5  X  150  =  375  Ibs.  Hence  the  supposed  beam 
rill  be  subjected  to  a  downward  acting  load, 

in  Case  (a),  of  2250  +  375  —    700  =  1925  Ibs.  sq.  ft., 
in  Case  (6),  of  3050  +  375  -  1550  =  1875  lbs-  sq-  ft 
The  first  case  gives  the  greater  load,  hence  substitute  w  = 
925,  /  =  10,  in  the  formula,  M  =  l/I2  wl2  X  12  =  1925  X  10* 


158 


DESIGN   OF   RETAINING   WALLS 


=  192,500  in.-lbs.     With  d  =  30  —  3  =  27",  we  have  by 
approximate  formula: 

M  192,500 


A  = 


=  0.510  sq.  in. 


fjd      ^ (16,000)  X  27 
Use  $/&"  D  bars,  8"  c  to  c,  giving  A  —  0.588  sq.  in.,  pla< 
3"  from  the  bottom  for   the  whole  length  and  $/Hr  D  bars, 
c  to  c,  placed  3"  from  the  top  at  counterforts  and  extehc 
2  ft.  beyond  them  on.  either  side. 

The  shear  at  the  edge  of  the  counterfort  is  Q  =  1925  X  4.: 
=  8180  Ibs. 

There  is  no  need  to  take  the  shear  at  the  center  of  counl 
forts,  since  there,  d  >  27". 

The  unit  shear  is,  taking  j  =  %  approximately, 
Q_  _          8180 
bjd~  12  Xj/sX  27 


=  29  Ibs.  sq.  in. 


For  the  s/8"  D,  8"  c  to  c,  2o  =  '}/  (2.5)  =  3.75 


U   = 


8/ 
=    % 


8l8° 


92.4  Ibs.  sq.  in. 


27  X  3-75 

For  similar  beams  nearer  the  face  wall,  the  soil  reaction 
greater  and  the  shear  less,  and  as  the  5/6"  D  bars,  8"  c  to 
will  be  used  throughout,  such  beams  are  over-reinforced  and 
can  assist  the  extreme  one  just  examined,  so  that  the  value  u  — 
92.4  Ibs.  sq.  in.  is  really  never  exerted.  In  fact,  the  heel  slab 
has  to  act  as  a  whole  and  is  materially  assisted  too,  by  its  con- 
nection with  the  face  wall  which  was  ignored  in  the  computation. 

Let  us  next  design  the  reinforcement  for  the  rearmost  2  ft. 
width  of  heel  slab,  including  the  projection,  Fig.  52.  This  gives 
a  beam  24"  wide,  54"  deep  and  10'  long  to  examine. 

It  is  easily  shown  that  in  Case  (a),  the  weight  of  beam  and 
earth  over  it,  less  the  soil  reaction,  is  56,000  Ibs.  nearly.  Hence 
proceeding  as  before, 

M  =  yi2  wL  I  X  12  =  56,000  X  10  =  560,000  in.-lbs. 


Hence  with  d  =  54  —  3  =  51" 
M        o  /      560,000 


A  = 


—    8/ 

~~    /7 


fajd         '  16,000  X  51 
This  is  the  area  of  metal  in  24"  width. 


0.785  sq.  in. 


:J  COUNTERFORTED  WALLS.       VERTICAL  RODS  159 

Use  three  $/&'  D  bars,  say  7"  c  to  c  .'.  A  =  1.176  for  three 
irs.  The  three  bars  are  to  be  placed,  3"  above  the  bottom 
iroughout.  Other  ^6"  Q  bars,  7"  c  to  c,  are  to  be  placed 
'  below  the  top  at  the  counterforts  and  will  extend  2'  6"  in 
ich  longitudinal  direction.  The  latter  will  provide  for  the 
3gative  moments. 

The  shear  at  edge  of  counterforts  =  l/2  (8.5  X  5600)  = 
5,800  Ibs.  =  Q. 

Unit  shear  =  v  =  %  —  =  %    23>  °°    =22.3  Ibs.  sq.  in. 
od          24  X  51 

For  bond  stress,  20  =  3  X  2.5  =  7.5,  for  the  3  bars, 


.-.  u  =  «/          =  y  _  =  ?1.8  ibs.  sq.  in. 

7  d  Vo      h  51  X  7-5 

Both  u  and  v  are  within  safe  limits. 

Vertical  Rods,  Tying  the  Heel  Slab  to  the  Counterfort.  In  the 
earn  i  ft.  wide,  2.5  ft.  deep  and  10  ft.  long,  regarded  as  cut 
at  of  the  heel  slab  next  the  projection,  the  load  on  it  between 
Dunterforts  was  found  above  to  be  1925  Ibs.  sq.  ft.  or  1925  X  8.5 
=  16,360  Ibs.  To  this,  add  the  weight  of  the  part  of  the  beam 
nder  the  counterfort,  1X1.5X2.5X150  =  560  Ibs.,  making 
total  of  16,920  Ibs.  as  the  load  to  be  carried  by  the  vertical 
ods.  This  requires  an  area  =  16,920/16,000  =  1.06  sq.  ins. 
i  12"  width. 

y^f  D  rods,  8"  c  to  c  will  be  placed  in  pairs  (giving  A  = 
.176  sq.  ins.)  or  rather  U  rods  will  be  used,  passing  under  the 
orizontal  rods  running  from  the  toe,  and  the  ends  will  extend 
ertically  upward  in  the  counterfort,  to,  or  near  to,  the  inclined 
ods.  As  the  load  on  the  vertical  ties  steadily  diminishes  as 
he  face  slab  is  approached,  $4"  D  rods  were  used  throughout, 
•ut  the  spacing  was  varied  as  indicated  in  Fig.  54.  Thus  there 
re  3  pairs  of  rods,  8"  c  to  c,  area  2.352,  and  nearer  the  face 
/all,  3  pairs  of  rods,  16"  c  to  c,  area  2.352,  giving  a  total  area 
>f  4.704  sq.  ins.,  that  can  carry  a  total  stress  of  4.704  X  16,000  = 
5,000  Ibs.  No  vertical  ties  are  needed  within  16"  from  the 
ace  slab,  as  it  is  compressive  area,  App.,  Art.  10. 


160  DESIGN  OF  RETAINING  WALLS 

As  a  check,  consider  the  whole  load  that  must  be  carried 
by  the  vertical  ties  to  the  left  of  the  ''projection." 

Weight  of  earth  over  heel  slab  between  counterforts,  Case  (a), 
=  8  X  8.5  X  22.5  X  ioo  =  153,000  Ibs. 

Weight  of  slab  10  ft.  long  =  8  X  10  X  2.5  X  150  = 
30,000  Ibs. 

Soil  reaction,  Case  (a),  =  ]/2  (2750  +  560)  X  8  X  10  = 
132,400  Ibs. 

Subtracting  this  from  the  sum  of  the  first  two,  we  have 
51,000  Ibs.  as  the  total  load  to  be  carried  by  the  vertical  rods. 
They  were  designed  to  care  for  a  total  stress  of  75,000  Ibs., 
which  is  more  than  adequate.  In  fact,  part  of  the  load  is 
carried  by  beam  action  directly  to  the  face  wall,  so  that  the 
total  is  really  less  than  51,000  Ibs.,  as  computed  above. 

Counterfort.  The  earth  thrust  on  the  face  wall  has  been 
supposed  to  be  carried  to  the  counterforts,  which  hypothesis  is 
on  the  side  of  safety,  since  part  of  this  thrust  below  and  slightly 
above  F  is  carried  directly  to  the  base. 

Assuming,  however,  that  the  horizontal  earth  thrust  on 
10  ft.  length  of  wall,  for  the  height  of  the  counterfort,  22.5  ft., 
is  all  carried  by  one  counterfort,  the  resultant  of  this  thrust  in 
amount  and  position  is  desired. 

In  amount  it  is,  see  Fig.  52, 

E  =  y*  (230  +  870)  X  22.5  X  10  =  123,750  Ibs. 
and  it  acts  above  C, 

/  h'      \   h        (  8        \22.5 

C  =    (  x  -\ )  —  =    (  !  -\ ) =   9.06  ft. 

V     r  h  +  2  h'J  3         V     r  22.5  +  !6/    3 

The  value  of  E  is  found  by  aid  of  the  pressure  diagram 
Fig.  52  (drawn  to  a  larger  scale),  where  the  eurth  pressures  at 
N  and  C  are  found  to  be  230  and  870  Ibs.  sq.  ft.  respectively. 
Similarly,  the  unit  pressures  at  /z  =  7,  h  =  12,  A  =  17,  are 
found  to  be  435,  580,  and  720  Ibs.  sq.  ft.  as  marked  on  the  figure, 
and  the  earth  thrusts  for  depths  of  7,  12,  and  17  ft.  are  computed, 
as  well  as  the  distances  c  from  the  lower  edge  of  the  area  pressed 
to  the  successive  resultants,  as  in  the  illustration  above.  The 
results  are  put  in  columns  E  and  c  of  the  following  table,  and 


COUNTERFORT 


161 


the  resulting  bending  moments  (in  ft.-lbs.)  are  then  computed 
and  inserted  in  the  column  headed  M  =  EC. 

The  reinforcement  of  the  counterfort,  for  purposes  of  com- 
putation, will  consist  of  inclined  rods  parallel  to  and  3"  from 
1C,  the  back  of  the  counterfort.  An  exact  computation,  in- 
cluding the  vertical  rods  that  tie  the  base  slab  to  the  counterfort, 
is  given  in  the  Appendix,  Art.  10  (also  see  Art  12),  but  it  will 
generally  suffice  in  practice  to  ignore  the  vertical  rods,  particu- 
larly as  it  is  on  the  side  of  safety. 

In  the  resulting  approximate  computation,  horizontal  sec- 
tions are  taken,  extending  through  the  counterfort  to  the  exposed 
face  of  the  vertical  slab.  For  any  section,  d  =  horizontal  dis- 
tance from  front  face  to  inclined  rods. 

By    App.,    eq.    (n),     Ms=fsA1j.d    cos    &=f3Aij.p, 

where   p  =  d  cos   ft  =  perpendicular   distance  from  the  front 

face  to  the  inclined  rods.     The  distances  p  from  points  on  the 

•front  face  at  /J=7,   12,   17,    22.5,    were   measured   to   scale 

jj  and  recorded  in  the  table.    Note  that  M  is  in  ft.-lbs.  and  p 

in  feet. 

Since  &  =  angle  that  1C  makes  with  the  vertical,  tan  fr  = 
10/22.5  =  °-445  •  •  ft  =  23°  58'-  Let  us  assume  a  steel  per- 
centage =  0.4;  then  from  App.  Fig.  9  (dotted  lines)  we  find 
j  =  0.906.  As  usual,  take  fs  =  16,000  .'.  the  formula  above 
reduces  to, 

12  M  =  16,000  X  0.906  X  12  p.  AI  .'.  AI  =  -      , 

14,500  p 

where  A i  is  in  square  inches.  The  areas  AI  are  those  of  right 
sections  of  the  inclined  rods  at  the  distances  h  ft.  below  the 
top  of  wall. 


& 

E, 
Lbs. 

c, 
Ft. 

M  =  EC, 
Ft.  Lbs. 

& 

4    •         M 

14500*' 

Sq.  In. 

7 

23,310 

3-15 

73400 

3-5 

1-45 

12 

48,600 

5-14 

249,800 

5-5 

3-14 

17 

80,750 

7-05 

569,300 

7-6 

5-i8 

22.5 

123,750 

9.06 

1,121,200 

9.8 

7.90 

162 


DESIGN   OF   RETAINING  WALLS 


Use  i/4"  D  bars,  two  extending  to  the  top,  two  extending 
to  8  ft.  from  the  top  and  finally  two  bars  to  extend  to  h  =  13, 
the  lower  sets  of  bars  being  extended  sufficiently  to  secure 
sufficient  bond  strength.  The  total  steel  area  of  the  six  bars 
=  9.45  sq.  in. 

For  Ai  =  9.45,  the  steel  ratio  =  9.45  -5-  (18  X  128)  =  0.0041, 
as  assumed.  The  results  are  all  in  excess,  which  justifies  the 
tacit  assumption  that  all  rods  are  in  the  same  inclined  plane, 
whereas  some  of  them  are  dropped  a  little  below  the  others, 

Fig.  54- 

The  lower  ends  of  the  rods  pass  into  the  projection  and  are 
hooked  over  the  lowest  bars  in  it,  giving  ample  bond  stress, 
since  with  hooked  ends  40  diameters  or  40  X  i)4  =  50"  is  all 
the  embedment  needed  and  51"  is  supplied. 

It  is  well  to  test  the  bond  and  shear  stresses.  In  doing  so, 
note  that  the  external  shear  =  E,  as  computed  above.  Also, 
the  breadth  of  the  counterfort,  b  =  18",  and  S0  =  10,  20  or 
30  sq.  ins.,  according  as  2,  4  or  6  bars  are  used.  Assume  j  =  0.906 
as  before. 

By  App.,  eqs.  (29)  and  (33),  neglecting  the  term  in  M,  we 

E  E 

have,  bond  stress  =  u\  = 


jd 


0.906 


unit  shear  =  v  =  — —  = 


E 


18  X  .906  d 
which  are  in  excess  of  the  true  values. 


16.3  X  d ' 


h 

E, 

d, 

Ml 

• 

Ft. 

Lbs. 

In. 

Lb.  Sq.  In. 

Lb.  Sq.  In. 

7 

23,310 

46 

56 

31 

12 

48,600 

72 

75 

42 

17 

80,750 

100 

45 

50 

22.5 

123,750 

128 

36 

59 

The  bond  stresses  are  within  working  limits,  as  indeed  are 
the  unit  shears,  since  the  cantilever  is  thoroughly  reinforced 
against  diagonal  tension,  not  only  by  the  inclined  rods,  but 


COUNTERFORT  163 

also  by  the  horizontal  and  vertical  bars,  which  act  as  stirrups, 
[n  such  cases,  shearing  stresses  of  from  60  to  120  Ibs.  sq.  in.  are 
illowed,  depending  on  the  amount  and  character  of  the  rein- 
"orcement  for  diagonal  tension. 

In  some  designs,  the  horizontal  and  vertical  bars  of  the 
:ounterfort  are  cut  off  as  soon  as  sufficient  embedment  is 
secured  to  carry  the  bond  stress,  but  it  seems  advisable  to 
extend  the  horizontal  bars  to  connection  with  the  inclined  ones, 
md  the  vertical  bars  the  full  length  to  the  inclined  bars,  or  at 
east  to,  say,  ^3  of  that  length.  Such  bars  act  like  stirrups 
n  resisting  diagonal  tension  and  are  very  effective  in  reinforcing 
:he  large  slab  against  shrinkage  stresses. 

Where  the  inclined  bars  are  bent,  the  radius  of  the  bend 
ror  square  bars  should  not  be  less  than  25  diameters,  and  for 
'ound  bars  20  diameters  (App.,  Art.  21),  otherwise  the  concrete 
.vill  be  over-stressed. 

Pull  of  Face  Slab  on  Counterfort.    The  earth  thrust  for  a  width 

8.5  ft.  between  the  counterforts  is  carried  by  beam  action 
}f  the  face  slab  to  the  counterforts.  The  corresponding  pull 
}f  the  vertical  slab  on  the  counterforts  is  resisted  by  the  horizon- 
:al  ties  of  the  counterfort,  above  alluded  to.  From  the  unit 
Dressure  diagram  of  Fig.  52,  the  total  earth  thrust  on  a  beam 
:ut  out  of  the  vertical  slab  i  ft.  high,  is, 

at  h  =    7,     435  X  8.5  =  3697  Ibs., 
at  h  =  22.$,  870  X  8.5  =  7395  Ibs. 

The  latter  thrust  will  require  a  metal  area  =  7395  -r  16,000 
=  0.462  sq.  in.,  which  can  be  supplied  by  a  pair  of  ^"  D  rods 
lying  in  the  same  horizontal  plane.  Such  pairs  are  spaced  12" 
apart  vertically  from  h  =  7  to  h  =  22.  From  the  top  of  the 
wall  to  h  =  7,  y%'  D  rods,  14"  apart  vertically  will  suffice. 

The  bond  stress  developed  in  the  12"  thickness  of  face  slab 
is  inadequate,  since  even  with  hooked  ends  the  length  of  em- 
bedment should  be  40  X  >£  =  20",  whereas  only  10"  is  available. 
It  is  plain  that  some  additional  means  of  resisting  the  pull 
must  be  provided.  One  simple  way  is  to  pass  the  horizontal 
rods  of  the  counterfort  around  or  through  vertical  steel  angles 


164  DESIGN   OF   RETAINING   WALLS 

placed  near  the  exposed  face  of  the  vertical  slab.  The  pairs 
of  rods  spoken  of  thus  become  continuous  U  rods. 

The  horizontal  bars  of  the  face  slab  can  either  pass  through 
holes  punched  in  these  angles  or  they  may  lie  entirely  outside 
the  angles.  The  effective  bearing  area  of  the  angles  should  be 
at  least  25  times  the  area  of  the  rods,  App.,  Art.  21. 

Reinforcement  against  shrinkage  and  temperature  stresses  in 
the  -vertical  slab.  To  reinforce  against  shrinkage  and  for  spacing, 
vertical  %"  D  bars,  24"  apart,  will  be  placed  near  the  front 
and  rear  faces. 

The  area  of  the  cross-section  of  the  vertical  slab  for  a  depth 
of  22  ft.,  is  22  X  12  X  12  =  3168  sq.  in.  The  metal  area 
required  for  temperature  reinforcement  is  y$  of  i%  =  10.56 
sq.  in. 

To  avoid,  as  much  as  possible,  a  multiplicity  of  sizes,  change 
the  48  horizontal  y^"  D  bars,  designed  for  beam  action,  to 
«M$"  D  bars,  similarly  spaced,  placed  near  the  front  face  and 
add  48  horizontal  y%'  D  bars,  similarly  spaced,  near  the  rear 
face,  as  shown  in  Fig.  54.  The  total  steel  area  is  48  X  0.392 
+  48  X  0.141  =  25.584  sq.  in.,  which  is  greater  than  the  total 
area  of  the  originals  yi"  D  bars  (=  12  sq.  in.)  +  10.56  sq.  in. 
required  for  temperatures  stresses. 

The  area  thus  added  for  temperature  stresses  will  be  found 
to  be  ample  for  h  =  7  to  h  =  22  and  sufficient  from  h  =  o  to 
h  =  7,  considering  that  the  steel  area  designed  for  beam  action 
over  this  upper  portion  is  considerably  in  excess. 

The  3/6"  D  bars  should  be  wired  at  intervals  to  the  l/2n 
D  bars,  5  ft.  each  in  length,  required  near  the  rear  face  to  resist 
the  negative  moments  at  the  counterforts. 

See  Fig.  54,  for  the  final  design. 

The  three  types  of  gravity,  T  and  counterforted  walls  hitherto 
discussed,  are  standard,  though  of  course  the  ratio  of  length  of 
toe  to  heel  varies  to  suit  special  conditions.  Thus  it  often 
happens,  on  account  of  a  limited  right  of  way,  as  in  track  eleva- 
tion in  cities,  that  the  toe  of  a  retaining  wall  must  be  made 
very  short  or  else  done  away  with  altogether,  the  resulting  type 
being  an  L  wall.  Some  interesting  types  of  walls  without  toes, 


COUNTERFORTED   WALLS 


165 


are  given  in  Engineering  News  for  April  20,  1911,  and  the 
issue  for  July  28,  1910,  shows  a  great  variety  of  types  of  gravity 
walls  as  actually  constructed 


All  plain 
Q  bars 


,. 


/ 

lensrth  7,'         , 

f  y^  \ 

\ 

L 


I 
I 

•^K--2-'-3H 
i  I 


FIG.  54 


In  Fig.  55*  is  shown  a  section  of  one  type  of  retaining  wall, 
ased  on  the  Chicago  Track-Elevation  work,  Rock  Island  lines. 

*  Given  in  Engineering  News,  April  8,  1915. 


166 


DESIGN   OF  RETAINING  WALLS 


The  walls  are  built  in  sections  35  ft.  long  to  allow  for  tempera- 
ture changes,  vertical  keys  being  used  at  the  expansion  joints. 
The  usual  weep-holes  are  omitted,  but  along  the  back  of  the 
wall  are  laid  inclined  drains  of  6-in.  porous  tile,  on  a  grade  of 

0.5%  extending  from  subgrade  level 
ultimately  to  an  8-in.  pipe  through 
the  wall,  having  connection  with  the 
catchbasin  of  a  city  sewer.  The  duct 
at  the  top  is  for  electric  wires,  cables 
and  telegraph  lines.  The  back  of  the 
wall  is  waterproofed  with  a  tar-pitch 
composition,  a  strip  of  burlap  and 
felt  being  placed  over  each  expansion 
joint  and  well  mopped  with  the  com- 
position. It  is  highly  desirable  to 
keep  water  out  of  the  expansion  joints, 
since  its  freezing  is  disintegrating  to 
the  concrete. 

The  reader   will  notice  the  usual 

stepped  wall,  which  is  always  advisable.  Sometimes,  in  cold 
climates,  the  inner  face  of  a  retaining  wall  for  3  or  4  ft.  down,  is 
battered  at  45°  or  less,  to  allow  for  the  freezing  of  the  earth 
and  its  consequent  expansion. 


Sidewalk 


El.6.25 


FIG.  55 


CHAPTER  V 


EARTH  ENDOWED  WITH  BOTH  COHESION  AND  FRICTION 

89.  THE  theory  will  now  be  developed  of  the  pressures  in 
an  unlimited  mass  of  earth  endowed  with  cohesion  as  well  as 
friction.  The  old  hypothesis  of  Coulomb  will  be  used,  that  the 
friction  on  a  plane  in  the  interior  of  a  mass  of  earth,  is  equal 
to  the  normal  pressure  multiplied  by  the  coefficient  of  friction, 
/  =  tan  <f>j  and  the  cohesion  is  equal  to  the  area  of  the  plane 
considered  multiplied  by  c,  the  cohesion  per  unit  of  area.  In 
other  words,  the  friction  is  proportional  to  the  normal  pressure 
on  the  area  considered,  whereas  the  cohesion  is  simply  propor- 
tional to  this  area  and  is  independent  of  the  normal  pressure. 

In  applying  the  following  theory,  it  is  essential  that  both 
<p  and  c,  be  determined  by  experiment,  one  method  for  experi- 
menting having  been  suggested  in  Art.  3.  It  would  be  erroneous 
to  assume  for  coherent  earth,  that  <p 
could  be  found  by  observing  the  natu- 
ral slope.  The  inclination  to  the  hori- 
zontal of  the  greatest  slope  at  which 
the  earth  will  stand,  is  always  greater 
than  the  true  angle  of  friction  <p  for 
coherent  earth.  This  is  a  consequence 
of  eq.  i,  Art.  3  and  it  will  be  inci- 
dentally shown  in  Art.  90,  where  the 
greatest  possible  surface  slope  is  found 
that  corresponds  to  a  given  angle  of 
internal  friction  <p  and  a  given  depth. 

In  Fig.  56,  ABC  represents  a 
wedge  of  rupture  at  a  depth  x  below 
the  free  surface,  whose  inclination  to 

the  horizontal  is  i.  The  wedge  is  supposed  to  have  a  length 
unity  perpendicular  to  the  plane  of  the  paper,  the  lengths  B  C, 
AB  and  AC,  being  taken  so  small,  that  the  unit  stresses  r,  r',  r" 

167 


FIG.  56 


168     EARTH  ENDOWED  WITH  BOTH  COHESION  JO)   FRICTION 

upon  the  faces  corresponding,  can  be  regardd  as  uniform.  As 
in  the  Rankine  theory  for  non-coherent  earth/,  the  unit  stress 
on  the  vertical  plane  AB  acts  parallel  to  the  op  slope  and  the 
vertical  unit  stress  r,  conjugate  to  /,  acts  on  a>lane  B  C  parallel 
to  the  top  slope  or  free  surface.  The  stresses  and  r',  thus  have 
a  common  obliquity  i. 

The  reaction  of  the  earth  below  the  planmf  rupture  AC  on 
that  plane  is  indicated  for  the  case  of  active  trust,  referring  to 
incipient  motion  down  the  plane.  In  this  ase,  the  normal 
unit  stress  being  />",  the  tangential  componen.  acting  up  along 
AC,  equals  q"  =  c  +  p"  f  =  c  +  p"  tan  &  y  Coulomb's  law 
quoted. 

The  frictional  unit  resistance  being  FE  =  'f  tan  <p,  it  follows 
that  EDF  =  tf>.  The  unit  cohesive  resistace  =  GF  =  c,  so 
that  the  total  tangential  unit  stress  q"  =  Gl  =  c  +  p"  tan  <p 
and  the  resultant  unit  reaction  =  GD  =  r" '.  The  obliquity  of 
stress  on  the  plane  AC  will  be  designated  hr  5;  .".  5  =  angle 
EDG.  We  have, 

GE       q"      c  +  p"tan<p  c 

tan  6  = =  —  =  -  -  =  /<; 

DE      p"  p"  p" 

The  obliquity  d  thus  increases  as  ^"  dimiishes  and  is  not 
invariable  as  for  non-coherent  earth,  where  $=  <p. 

The  wedge  of  rupture  with  the  unit  streses  acting  on  its 
faces,  as  drawn  here,  corresponds  to  Case  (b)  .Fig.  31,  when  c 
is  made  zero.  Let  the  student  draw  figures  fc  the  other  three 
03568  (<0»  (c)  and  (<f),  only  including  the  cohean;  remembering 
that  in  the  case  of  passive  thrust,  (c)  and  (d),  he  external  force 
r7  is  on  the  point  of  moving  the  wedge  AB(  in  the  direction 
A  to  C  and  q"  =  c  +  p"  tan  <p  now  acts  in  dizction  C  to  A  or 
opposed  to  the  motion. 

The  circular  diagram  of  stress  is  shown  i  Fig.  57. 
ordinate  OL  ±  KOJ,  is  laid  off  equal  to  c  anc  the  line  KLS  is 
drawn  through  L,  making  the  angle  <p  with  le  line  KOJ ';  its 
equation  being,  when  p"  and  q"  are  the  variably 

<f*  =  c  H-  p"  tan  <p 
p"  and  .".  q",  varies  with  the  depth  .v. 


ERCULAR    DIAGRAM   OF    STRESS 


169 


We  have  inn  the  figure,  KO  =  c  cot  <p.  To  draw  the 
circle,  for  the  ae  of  active  pressure,  lay  off  OR  =  r  =  wx  cos 
i  =  wh,  makingthe  angle  ROJ  =  i  with  OJ  and  then  find  by 
trial,  the  centerP  on  OJ,  so  that  the  circle,  with  center  P  and 
the  least  radius  ..hall  pass  through  R  and  be  tangent  to  KLS. 
Let  the  tangen  point  be  written  S.  Then  if  p"  =  ON,  we 
have,  f  =  c  +>"  tan  <?  =  NS  and  tan  NOS  =  tan  d  =  $"/#", 
as  should  be  for. he  plane  of  rupture  AC  of  Fig.  56. 

Xow  IS,  Fii  57,  corresponds  to  this  plane  of  rupture  AC, 
Fig.  56,  since  foany  other  plane  through  /  as  IR,  the  tangential 


FIG.  57 

component  M Tftvould  be  less  than  the  corresponding  ordinate 
<{'  to  KS  (proaced  if  necessary)?  so  that  slipping  could  not 
be  impending  a  the  plane  corresponding  to  IR.  Slipping  is 
thus  impending)nly  on  the  plane  corresponding  to  IS,  so  that 
it  is  one  planeof  rupture.  We  cannot  determine  the  point 
5  as  in  Art.  6 2  since  the  obliquity  8  varies  with  p"  or  with  x 

/  c  -  p"  tan  <p\ 

( for,  tan  d  = ]  and  in  fact,  5  is  unknown  until  ascer- 

P  ' 

tained  by  this  instruction.    The  circle  drawn  is  the  only  one 

that  will  satisf  the  equations  above  for  the  plane  of  rupture.* 

The  relatioroetween  Fig.  57  and  the  ellipse  of  stress  Fig.  29, 

*This  diagrams  similar  to  the  one  given  by  Prof.  O.  H.  Basquin  for 
coherent  earth,  irhis  paper  on  "The  Circular  Diagram  of  Stress  and  its 
Application  to  th  Theory  of  Internal  Friction,"  Western  Soc.  of  Engs., 
Sept.  9,  1912. 


170     EARTH  ENDOWED   WITH   BOTH   COHESION   AND   FRICTION 

should  be  kept  clearly  in  mind.     Thus,  if  OR,  Fig.  57,  cuts  the 
circle  in  R' ',  then  if  we  assume 

OP  =  OP'   of  Fig.  29  =  OP  of  Fig.  57 
PR  =  P'R'  of  Fig.  29  =  PR  =  PR'  of  Fig.  57, 
also  that  the  obliquity,   ROP  =  R'OP' ',  is  the  same  in  both 
figures,  then  it  follows  that  OR  in  one  figure  =  OR  in  the  other, 
and  similarly  for  OR'.     Hence  the  lengths  OR,  OR',  of  Fig.  57 
represent  the  conjugate  stresses  OR,  ORf  of  Fig.  29,  having  the 
common  obliquity  ROP.    Also  01  and  OJ  give  the  lengths  of 
the  semi-axes  of  the  ellipse  of  stress. 

If  the  student  will  now  re-read  Arts.  63-64  and  the  part  of 
Art.  65  not  involving  the  derivation  of  (41)  and  (42),  he  will 
find  all  the  conclusions  to  hold  on  simply  replacing  the  p  of 
Fig.  29  by  the  corresponding  5  of  Fig.  57.  The  latter  figure 
is  thus  the  true  circular  diagram  of  stress.* 

To  find  the  active  conjugate  thrust  rr  for  a  given  vertical 
depth  x,  we  compute  r  =  wx  cos  i  =  wh  and  having  laid  off 
the  angle  ROJ  =  i  and  OR  =  wh  (to  the  scale  of  force)  we 
draw  the  circle  of  least  radius  that  passes  through  R,  touches  .KL  S 
and  has  its  center  on  KOJ.  If  this  circle  cuts  OR  at  R',  then 
OR',  to  the  scale  of  force  =  /,  Art.  63. 

To  find  the  passive  conjugate  stress  r'  for  the  given  vertical 
depth  x,  or  the  normal  depth  h,  lay  off,  to  the  scale  of  force, 
r  =  wh  =  OR'  and  draw  the  circle  of  largest  radius  through  R', 
tangent  to  KLS  and  with  its  center  on  KOJ.  Then  if  OR'  pro-, 
duced  cuts  the  circle  at  R,  the  conjugate  passive  thrust  =  /  = 
OR,  to  the  scale  of  force.  Certain  conclusions  of  Art.  65  likewise 
apply  to  the  angles  0,  7,  /?',  y'  (having  the  relative  positions 
shown  in  Fig.  36)  that  the  planes  of  rupture  make  with  the 
vertical.  Thus  in  Fig.  57,  ft  =  ^  SPR,  /  =  SPI  -  ft  = 
90°  -  9  -  ft  ft  =  y2  SPR,  y'  =  SPJ  -  ft  =  90°  +  <p  -  ft. 
Compare  with  Fig.  36,  noting  that  SOP  is  now  =  5.  When 
c  =  o,  then  d  =  <p  and  the  construction  of  Fig.  37  for  non- 
coherent earth  applies.  For  i  <_  <p  and  x  very  large,  c/OR  is 
very  small  and  the  results  approach  those  for  c  =  o. 

*  It  will  prove  instructive  to  the  student  to  draw  the  ellipse  of  stress  in 
true  position,  after  the  method  outlined  in  "Note,"  Art.  63. 


19,  90]                               MAXIMUM  SURFACE    SLOPE                                    171 

EXAMPLE.     Let  c  =  100  Ibs.  per  sq.  ft.,  w  =  100  Ibs.  per  cu.  ft.,  <f>  =  i  = 

o°,  find  by  the  construction  for  active  thrust,  the  values  of  the  angles  y, 

nade  by  the  plane  of  rupture  IS  with  the  vertical  R\I  and  the  values  of  the 

on  jugate  stress  r'  for  the  values  of  x  given.* 

X 

Ft. 

h 
Ft. 

7 

r' 

Lb.  per  Sq.  Ft. 

0 

3-46 

3-00 

30° 

0 

30° 

5.26 

4-55 

35° 

84 

25° 

8.42 

7.29 

40° 

253 

20° 

15.00 

13.00 

45° 

646 

15° 

32.70 

28.30 

50° 

1,847 

10° 

124.00 

107.30 

55° 

8,738 

5° 

If  the  values  of  x  are  laid  off  along  a  vertical  axis  and  the  values  of  r' 
ilotted  as  horizontal  ordinates,  it  will  be  found  that  the  line  joining  the 
:xtremities  of  the  r"s  is  very  slightly  concave  upwards.  The  stress  is  thus 
lot  uniformly  increasing  and  the  center  of  pressure  is  very  slightly  below  the 
:orresponding  center  for  a  stress  uniformly  increasing  from  the  depth  x  =  3.46 
t.,  where  r'  =  o  to  the  full  depth  considered. 

If  a  line  be  drawn  parallel  to  the  surface,  at  a  depth  3.46  ft.  below  it 
;  where  r  '  =  o)  and  this  be  regarded  as  a  new  surface  for  non-coherent  earth, 
;he  unit  thrust  for  x  =  124  ft.  or  for  the  depth  (124  —  3.5)  referred  to  the 
lew  surface  for  non-coherent  earth  is,  by  the  usual  formula  for  such  earth, 
•120.5  cos  <p  =  10,440  Ibs.  per  sq.  ft.  This  differs  materially  from  the  true 
|:hrust  8738  Ibs.  found  above.  See  a  corresponding  comparison  in  Art.  96, 
"or  earth  surface  horizontal. 

It  is  seen  from  the  table  that  y  increases  with  x,  hence  the  surface  of  rup- 
ture considered  makes  an  angle  with  the  vertical  which  increases  with  the 
depth.  Hence  the  entire  surface  of  rupture  from  x  =  3.46  ft.  down,  is  concave 
upwards.  For  x  =  3.46,  r'  =  o  or  the  points  O  and  I  coincide.  For  x  <  3.46 
ft.,  it  will  be  shown  in  Art.  92,  for  a  stable  unlimited  mass  of  earth,  that  the 
conjugate  stress  r'  =  o. 
: 

90.  Maximum  Surface  Slope  for  a  Given  Depth. 


<  i  <  45  °  H 


It  is  evident  from  Fig.  57,  for  active  thrust,  where  r'  <  r,  since 
the  circle  drawn,  passing  through  R,  center  on  OJ  and  touching 
KLS,  is  the  one  with  the  smallest  radius,  that  OR  cannot  lie 
to  the  left  of  05;  hence  for  active  thrust,  the  greatest  value, 


*  These  values  were  in  reality  computed  from  formulas,  taking  y  for  the 
independent  variable.     They  were  checked  by  the  diagram. 


172    EARTH  ENDOWED  WITH  BOTH   COHESION  AND   FRICTION 

for  a  given  depth  •#,  that  i  can  attain  is  6,  when  OR  coincides 
with  05.*  A  simple  construction  will  give  this  maximum  value 
of  i,  when  c,  <p  and  the  normal  depth  h  =  x  cos  i  are  given. 
Thus  in  Fig.  58,  having  laid  off  OL  =  cy  where  OL  is  J_  KJ, 
and  drawn  the  line  KLS  making  the  angle  <p  with  KJ\  then 
since  i  =  d,  we  have,  r  =  wx  cos  i  =  wh  =  OS.  Hence,  with 
O  as  center,  wh  (to  scale)  as  radius,  describe  an  arc,  cutting 
KL  produced  at  5;  whence  Z  SOJ  =  maximum  value  of  i  for 
the  normal  depth  h  assumed.  After  i  has  been  found  by  this 


FIG.  58 


FIG.  59 


construction,  then  x  =  h  sec  i,  can  be  computed,  but  x  cannot 
be  assumed  in  the  first  instance. 

As  a  special  case,  when  the  point  O  coincides  with  7,  Fig.  59, 
or  for  a  normal  depth  h  such  that  wh  —  OS  =75,  then  since 
KPS  =  90°  —  <p  and  from  the  isosceles  triangle  OPS,  2  SOP  = 
1 80°  —  OPS  =  90°  +  <p  .'.  the  maximum  value  of  i  =  SOP 

=  SIJ  =  4S°  +  -- 

2 

At  this  depth,  the  conjugate  stress  r1  =  OR'  =  o.  See  Art. 
92  for  the  corresponding  value  of  x.  As  the  vertical  depth  in- 
creases beyond  this,  the  length  05,  Fig.  58,  increases,  hence 
the  maximum  value  of  i  =  SOJ  diminishes  and  approaches  the 
angle  <p  for  great  depths,  since  then  Z  O5L  is  very  small.  For 

*  For  passive  thrust,  where  the  largest  radius  is  used  to  describe  a  circle 
with  center  on  0  J,  passing  through  Rr  and  touching  KLS,  OR'R  can  lie  above 
OS  and  it  is  possible  to  have  i  >  $.  This  can  be  visualized,  if  in  Fig.  57,  0  is 
taken  near  J,  so  that  OR'R  lies  above  OS.  In  this  case,  r'  =  OR,  can  be  less 
than,  equal  to  or  greater  than  r  =  OR'. 


0,91]  GREATEST   DEPTH  WHEN  i   >  <fi  173 

oherent  earth,  maximum  i  =  SOJ,  Fig.  58,  is  always  greater 
ban  <p. 

Having  found  by  the  preceding  construction  the  greatest 
>ossible  surface  slope  for  a  given  h,  if  for  this  i,  h  is  further 
acreased,  slipping  will  occur  at  the  new  depth,  unless  the  earth 
,t  that  depth  is  confined  by  walls,  natural  or  otherwise,  which 
prevent  motion  and  whose  resistance  thus  introduces  extrane- 
•us  forces  not  contemplated  in  the  theory  of  the  homogeneous 
nass  of  earth  of  indefinite  extent,  subjected  to  no  external 
orce  but  its  own  weight. 

For  the  limiting  case  above,  i  =  8  or  when  the  surface  attains 
ts  greatest  possible  slope  for  the  given  depth,  the  construction, 
rig.  58,  gives  the  inclination  of  the  planes  of  rupture  to  the 
vertical,  at  the  given  depth. 

As  proved  in  Art.  65,  when  the  figure  58  is  supposed  to  be 
evolved  about  /  until  the  chord  IRi  is  vertical  (or  RJ  horizon- 
;al),  then  for  active  thrust,  the  planes  of  rupture  IT  and  75 
vill  have  their  true  directions.  The  plane  IT  makes  the  angle 
RJT  with  the  vertical.  This  angle  =  RfIS,  is  measured  by 
'/2  arc  R'S,  which  is  also  the  measure  of  the  angle  OSK  =  i  —  v, 
vhence  /3  =  RiIT  =  i  —  <p.  By  Art.  65,  this  corresponds  to 
:ase  (a). 

For  case  (b),  shown  in  Fig.  56,  the  plane  of  rupture  correspond- 
ng  to  AC,  is  IS  in  Fig.  58,  which  makes  an  angle  with  the 
lorizontal  R\J  which  is  measured  by  %  (SJ  —  Ril)  =  y£ 
[SJ  —  R'l),  which  is  the  measure  of  d  =  i.  Hence  the  second 
Diane  of  rupture  is  parallel  to  the  free  surface.  See  Art.  94, 
Fig.  61,  for  application  of  the  conclusions  above. 

91.  The  Greatest  Depth  at  which  Equilibrium  is  Possible 
ivhen  i  >  <p.  When  the  surface  slopes  indefinitely,  its  angle  with 
the  horizontal  being  i,  it  has  been  seen  that  for  i  =  8,  slipping 
is  impending  on  the  plane  AC,  Fig.  56  where  r  =  wh  =  OS, 

Fig.  58- 

From  the  triangle  OLS,  Fig.  58,  we  have, 

r        OS      sin  OL  S       sin  (90°  +  <p)  cos  (p 


OL      sin  OSL      sin  (  i  —  <p)      sin  (i  —  <f>) 


174     EARTH   ENDOWED   WITH   BOTH   COHESION   AND   FRICTION 

Letting  x0  be   the    corresponding   vertical    depth   .'.    r 
w  x0  cos  i, 

r  c  cos  <p 

* 


i  > 


<P. 


w  cos  i      w  cos  i  sin  (i  —  <p) ' 

When  the  surface  slopes  indefinitely  as  assumed,  equilibrium 
is  impossible  at  a  greater  depth  than  x0  for  given  values  of 
i  and  <p. 

When  i  reaches  the  value  45°  H ,  the  formula  reduces  to 

2 

2  C  /  <p  \ 

x0  =  —  tan  (  45°  +  — ),  which  is  in  agreement  with  a  result 

1!J  \  9  / 


above,  when  the  points  O  and  /  coincide.     See  Art.  92. 

When  i  <  <p,  it  is  seen  from  Fig.  57,  that  we  cannot  have 
i  =  5;  in  this  case  there  can  be  equilibrium  at  any  depth. 

To  derive  a  formula  for  the  active  conjugate  stress,  r'  =  OR', 

C  COS  (p 

corresponding  to   the  depth  xot   or  to  r  =  OS 


sin  (i  — 
note  in  Fig.  58,  that, 

OPR'  =  IPS  -  R'PS  =  IPS  -  2.0SK  = 

(90°  —  <p)  —  2  (i  —  <p)  =  90°  —  (2  i  —  <p) 

Now  apply  the  law  of  sines  to  triangles  IPS  and  OR' P  in 
turn  and  substitute  the  above  value  of  r. 

sin  i  c  sin  i 

PS  =  OS 


cos  <p       sin  (i  — 

=  QR>   =   pR,  ~ 


sin  i  sin  (i  —  <p) 

where  r0'  is  the  active  conjugate  stress  at  the  depth  x0. 
As  seen  in  Art.  90,  at  this  depth, 


i  —  (p,  7  =  90°  —  i. 


92.  The  vertical  depth  x',  at  which  the  conjugate  stress  r;  is 
zero,  can  be  easily  found.     In  this  case,  O  coincides  with  7, 


)2,93J         DEPTH  AT  WHICH  CONJUGATE   STRESS   IS  ZERO  175 

lince  then  r'  =  OR'  =  o}  ROJ  =  i  and  in  Fig.  59,  since  KPS= 

p°  —  <p,  we  have  OPL  =  45° ;   whence  OP  =  OL  cot 

2 
f45°  -  — )  =  c  tan  (45°  +  — ) .     Noting  that  ORJ  =  90°,  OR 

\  2  /  2  ' 

r  =  2  OP  cos  i  =  2  c  cos  i  tan  (45°  -f  -  —  J . 


Also,  since  r  =  wx'  cos  i, 


This  important  formula  gives  the  vertical  depth  xf  at  which 
here  is  no  active  thrust.     It  is  true  for  any  i  not  greater  than 

i  =  SOJ  =  45°  +  —  • 

2 

From  Fig.  59,  it  is  seen  that  when  /  =  OR'  =  ORi  =  o, 
RI  coincides  with  O;  hence  the  horizontal  R\J  or  OJ  is  now  in 

xue  position.     Z  SOL  =  Z  LPO  =  45°  —  —  and  the  planes  of 

•upture  OS  and  OT  make  angles  of  (45°  —  —  )  with  the  vertical. 

^  2  / 


FIG.  60 


93.  Typical  forms  of  the  surfaces  of  rupture,  when  i  <  <p, 
are  shown  in  Fig.  60.    The  curves  were  sketched  for  <p  =  30°, 


176    EARTH  ENDOWED   WITH  BOTH  COHESION  AND   FRICTION 

i  —  15°,  c  =  100  Ibs.  sq.  ft.,  w  =  100  Ibs.  cu.  ft.     From  Art.  92, 

2  C  /  <p  \ 

we  have  x'  —  — tan  (45° )  =  2  tan  60°  =  3.46  ft. 

W  V  2  ' 

For  ac/W£  thrust,  the  surfaces  of  rupture  begin  at  the  depth 

x',  say  at  A,  where  they  make  angles  of  (45° )  =  30°  on 

v  2  / 

either  side  of  the  vertical,  Art.  92.  The  inclination  for  any 
depth  is  readily  found  from  the  circular  diagram  of  stress,  so 
that  the  (dotted)  curves  of  rupture  can  be  approximately  drawn.* 


h', 
Ft. 

X, 

Ft. 

ft 

7 

Lbs.  per  Sq.  Ft. 

3-33 

3-46 

30.0° 

30.0° 

0 

10 

10-35 

25-6 

34-4 

250 

20 

20.70 

23-9 

36.i 

640 

30 

31-05 

23-3 

36.7 

1,020 

40 

41.40 

23.0 

37-o 

I,4OO 

50 

51-75 

22.9 

37-1 

1,790 

500 

517.50 

22.0 

38.0 

19,200 

It  will  be  noticed  that  r'  varies,  very  nearly,  according  to  a  linear  law. 

One  way  to  do  this  is  to  draw  lines  parallel  to  the  free  surface 
at  the  normal  depths  h  at  which  the  inclinations  of  the  surface 
of  rupture  have  been  found.  At  A  draw  a  line  inclined  30° 
to  the  vertical.  At  some  point  on  this  tangent,  correspond- 
ing to  a  normal  depth  intermediate  between  the  first  and  second 
normal  depths,  draw  a  line  having  the  inclination  of  the  plane 
of  rupture,  corresponding  to  the  second  normal  depth  and 
continue  it  to  and  beyond  intersection  with  the  line  drawn 
parallel  to  the  surface  at  this  second  normal  depth.  It  repre- 
sents the  tangent  to  the  curve  at  the  intersection  point. 

Continue  thus  until  a  sufficient  number  of  enveloping  tangents 
are  drawn  to  sketch  the  curve.  Where  the  successive  tangents 
make  angles  of  only  2°  or  3°  with  each  other,  the  results  are  close 

*  In  the  table  following,  the  values  of  /?,  7  =  (90°  -  tp)  -  /?  =  (60°  -  /?) 
and  the  active  conjugate  thrust  r',  are  given  for  various  values  t)f  h  (assumed) 
and  x  =  h  sec.  i  =  1.035  h,  taking  c  =  100  lb./ft.2,  w  =  100  lb./ft.3,  v  =  30°, 
*  =  15°. 


93]  SURFACES  OF  RUPTURE  177 

approximations.  As  hitherto  pointed  out,  for  great  depths, 
the  slope  of  the  curve  AS  or  A  T  approaches  that  of  the  surface 
of  rupture  IS  or  IT  for  non-coherent  earth. 

Similarly,  the  curves  for  passive  resistance  (to  be  discussed 
later)  corresponding  to  motion  up  the  plane  A'T',  A'S',  at  con- 
siderable depths,  approach  in  inclination  the  surfaces  of  rupture 
I'T'  and  I'S'  for  non-coherent  earth. 

It  will  be  observed  that  the  surfaces  of  rupture  AS  and 
AT  are  nearly  plane  for  i  =  15°,  <p  =  30°,  assumed.  As  i 
increases,  the  curvature  becomes  more  pronounced.  When 
i  =  <p,  it  has  been  previously  pointed  out,  Art.  61,  that  75  is 
parallel  to  the  free  surface  and  IT  is  vertical.  Hence  f or  i  =  <p 
at  great  depths,  the  curve  AS  approaches  parallelism  to  the 
surface  and  A  T  approaches  the  vertical. 

Similarly,  when  i  =  <p,  I'S'  is  parallel  to  the  free  surface  and 
I'T'  becomes  vertical.  Hence  A'S' ,  as  x  increases  indefinitely, 
approaches  parallelism  to  the  free  surface  and  A'T'  approaches 
a  vertical. 

Recurring  to  the  numerical  example  of  Art.  89,  it  may  be 
of  some  interest  to  know  that  the  equation  of  the  curve  of 
rupture,  AS,  for  active  thrust,  was  found  to  be,  very  nearly,* 

y  =  0.2936  a:1-63 

It  must  be  borne  in  mind  that  this  equation  only  refers  to  the 
special  values,  i  =  <p  =  30°,  c  =  w  =  100,  the  greatest  depth 
entering  into  the  computation  being  124  ft.  However,  the 
curve  approaches  parallelism  to  the  free  surface,  as  x  increases 
indefinitely,  as  should  be  the  case.  Here  y  is  measured  along 
the  surface,  say  from  A'  to  the  left,  and  x,  as  hitherto,  is  measured 
vertically  downward.  The  curve  is,  of  course,  limited  to  values 

Referring  again  to  Fig.  60,  it  is  well  to  recall  that,  in  the 
unlimited  mass  of  earth  supposed,  the  conjugate  thrusts  on  a 
vertical  plane  act  parallel  to  the  free  surface.  To  appreciate 
the  significance  of  the  curves,  it  may  be  noted,  relative  to  A'T', 


*  The  derivation  of  this  equation,  which  involves  a  modification  of  a 
well-known  application  of  the  method  of  least  squares,  is  omitted,  since  it  is 
mainly  of  academic  interest. 


178     EARTH  ENDOWED   WITH  BOTH   COHESION  AND   FRICTION 


that  if  an  external  force  acts  parallel  to  the  surface  and  upward, 
sufficiently  great  to  cause  motion,  the  mass  to  the  left  of  A'T' 
will  move  up  along  the  surface  of  rupture  T'A'.  If  the  mass 
above  A  fSr  is  subjected  to  a  force  acting  down  and  parallel  to 
the  free  surface,  sufficiently  great  to  cause  motion,  the  mass 
will  move  up  along  the  surface  of  rupture  S' A' .  For  the  active 
thrusts,  if  the  mass  of  earth  to  the  left  of  AS  gives  way  from 
any  cause,  it  will  descend  along  the  surface  of  rupture  AS,  caus- 
ing, too,  a  break  above  A.  Lastly,  if  the  earth  to  the  right  of 
AT  gives  way,  it  will  descend  along  the  surface  of  rupture 
A  T,  the  break  above  A  following  as  a  consequence. 

94.  Similar  remarks  apply  to  Fig.   61,  where  i  is  greater 

than  <p  but  less  than  45°  +  — ,  Art.  90.    Here  AT,  A'S',  are  the 

2 

curves  of  rupture  for  passive  thrust,  Art.  89,  and  AS  and  AT 


FIG.  6 1 


are  the  curves  for  active  thrust.     To  sketch  the  latter,  we  first 
lay  off  vertically  downward  from  the  free  surtace, 

2  C  f  (D\  C  COS  <p 

A' A  =  —  tan  (45°  +  -),  BC  =  -          .    . -, 

w          *  2'  w  cos  i  sin  (i  —  <p) 

Arts.  91,  92,  and  through  the  point  C,  draw  a  parallel  S'C  TT' 
to  the  surface.     At  A,  the  curves  AS  and  AT  make  angles  of 

(45° J ,  on  either  side  of  the  vertical,  with  the  latter,  Art.  92. 


94] 


SURFACES  OF  RUPTURE 


179 


The  curve  AS  is  concave  upward  and  approaches  CT  produced 
as  an  asymptote,  Art.  90.  The  curve  AT  is  concave  down- 
ward and  at  T  makes  the  angle  (i  —  <p)  with  the  vertical, 
Art.  90.*  The  general  shape  of  the  curves  can  now  be  sketched. f 


h, 
Ft. 

X, 

Ft. 

r'  , 
Lb.  per  Sq.  Ft. 

0 

7 

5-18 

5-71 

O 

35°  oo' 

35°  oo' 

10.00 

11.03 

370 

24°  45' 

45°  15' 

15.00 

16.54 

870 

17°  15' 

52°  45' 

20.00 

22.06 

1,550 

10°  15' 

59°  45' 

21.60 

23.80 

1,990 

5°  oo' 

65°  oo' 

It  is  seen  that  r'  does  not  vary  uniformly  with  x,  the  rate  increasing  as  x 
increases.  This  might  have  been  anticipated  because  the  surface  of  rupture 
AS  is  so  far  removed  from  a  straight  line.  Regarding  r'  as  varying  uni- 
formly between  the  points  given,  the  total  active  earth  thrust,  acting  parallel 
to  the  surface,  is  found  to  be  14,150  pounds.  Its  resultant  acts  below  the 
third  point. 

It  has  been  proved,  Arts.  91-92,  that  at  A  the  active  con- 


jugate stress  r '  =  o  and  at  T  or  S,  r'  = 


c  cos  (2  i  —  (p) 

sin  (i  —  <p) 


For 


a  greater  depth  than  BC,  equilibrium  is  impossible,  if  the  earth 
is  free  to  move  along  CT.  It  would  seem,  then,  if  a  deep  cut 
was  made  in  a  hillside  of  homogeneous  earth  below  the  line 
CT,  that  sliding  would  occur  along  such  a  surface  as  AS. 

When  i<  45°+-  it  will  be  found  that  A  A'  <  BC,  but 


*r  1 

when  i  attains  its  limit,  45°  -\ Art.  90.  then  A  A'  —  BC,  and 

2 

equilibrium  is  not  possible  at  a  greater  depth  than  A  A',  unless 


*  The  table  gives  the  values  of  r'  (active  thrust),  /?  and  y,  correspond- 
ing to  various  values  of  h  (or,  x  =  h  sec.  i),  when  c  =  100  lb./ft.2,  w  =  100 
lb./ft.3,  <p  =  20°  and  i  =  25°. 

t  Figs.  60,  61,  are  similar  to  those  given  by  Resal  in  Poussee  des  Terres, 
II.,  pp.  34  and  38. 


180     EARTH  ENDOWED   WITH   BOTH   COHESION  AND  FRICTION 

some   external   forces   are  introduced,   not   considered   in   the 
theory  above.* 

95.  PASSIVE  THRUST.  In  Fig.  57  the  vertical  unit  stress  on  a  plane 
parallel  to  the  surface,  at  the  depth  x,  r  =  wx  cos  i  =  wh,  is  now  represented 
by  OR'  and  the  conjugate  passive  unit  thrust  by  OR.  The  angles  with  the 
vertical  made  by  the  planes  of  rupture  are, 

0'  =   y2   SPR,  7'  =  SPJ  -  ft'  =  90°  +  9  -  ft'. 

At  the  surface  where  r  =  OR'  =  o,  O  is  at  I,  Fig.  59,  and  as  proved  in 
Art.  92, 


r    =  2  c  cos  i  tan 

\  " 

•ft'  =  ^  SPR  =  SIR  =  45°  +  — -  i, 

y'  =  SPJ  -  ft'  =  90  +  <?  -  ft'  =  45°  +  —  X  *. 

These  are  the  angles  made  by  the  curves  A'T',  A'S',  Fig.  61,  with  the  vertical 
at  the  point  A'.^ 

When  i  >  ip  and  5  =  i,  Fig.  58,  since  now  r  =  wh  =  OR',  r'  =  OS,  we 
have  from  the  results  of  Art.  91, 

r'  =  OS 


sin  (i  —  </))' 

c  cos  (2  i  -  <p) 

r  =  OR    =  —  — —  =  wx  cos  ^, 

sin  (i  —  (/?) 

or  r  and  r'  for  active  thrust  are  interchanged  for  passive  thrust. 

Hence  at  the  depth  x  given  by  the  last  formula  ft'  =  %  SPR  =  o,  y'  = 
SPJ  =  90°  +  <p\  .'.  at  this  depth,  the  curve  A'T',  Fig.  61,  is  vertical  and 
the  curve  A'S'  is  inclined  at  the  angle  90°  +  <f>  with  the  vertical. 

*  If  in  Fig.  61,  we  take  a  plane  rock  surface  parallel  to  the  earth  surface, 
at  the  vertical  depth  x,  the  unit,  pressure  on  this  plane  is  r  =  wx  cos  i  = 
weight  of  vertical  prism  of  earth  resting  on  unit  area.  The  components  of  r 
parallel  and  normal  to  the  plane  are  r  sin  i,  r  cos  i,  respectively.  For  stable 
equilibrium,  we  must  have,  assuming  c  and  tan  <f>  the  same  for  earth  on  rock 
as  for  earth  on  earth, 

r  sin  i  ^.r  cos  i  tan  <f>  -J-  c, 
or, 

<  c  cos  V 

*  ~  w  cos  i  sin  (i  —  <p) 

The  rjght  member  is  the  value  of  BC,  previously  found,  so  that  it  is  again 
proved,  that  for  a  depth  greater  than  BC  equilibrium  is  impossible.  The 
result  refers  to  a  uniform  slope  of  earth  surface  and  rock  surface  of  indefinite 
extent,  never  exactly  realized  in  practice. 

t  The  horizontal  coordinates  of  curve  A  'Sf  are  very  much  reduced  to  keep 
the  figure  within  bounds.  The  true  curve  lies  above  the  one  shown  and  ex- 
tends much  farther  to  the  right. 


i]  PASSIVE    THRUST  181 

When  i  >  5,  as  noted  in  Art.  90,  OR'R  lies  above  OS  (consider  Fig.  57, 
hen  o  is  near  I)  and  ft'  becomes  negative.  With  R  at  S,  R'  is  on  the  arc 
RfS  below  5.  As  R'  moves  up  along  the  arc,  R  moves  down.  At  one 
>int  OR  =  OR'  is  tangent  to  the  arc  IS;  finally  when  R'  reaches  S,  R  has 
e  position  first  occupied  by  R'.  It  is  assumed  that  the  proper  circle  is 
awn  for  each  intermediate  position. 

When  R'  is  at  S,  r  =  wh  =  wx  cos  i  =  OS,  or  at  the  depth  x  =  x0,  Art.  91, 
e  conjugate  passive  thrust  r'  =  OR  =  r0'  =  the  active  conjugate  thrust, 
.e  computation  of  Art.  91,  exactly  applying,  considering  that  for  passive 
rust  r  =  OR'  =  OS,  r'  =  OR,  so  that  R  and  R'  are  interchanged.  We 
ive  also  ft'  =  —  (*'  —  <p),  y  =  90  -f-  i. 

Since  the  active  and  passive  thrusts  are  equal  at  the  depth  x0  the 
tiuilibrium  there  is  unstable. 

EXAMPLE.  Let  i  =  25°,  <p  =  20°,  c  =  w  =  100.  From  the  results  above, 
e  passive  thrust  at  the  surface  =  2  c  cos  25°  tan  55°  =  259  Ibs.  per  sq.  ft. 

.id  ft'  =  30°,  7'  =  80°.     At  the  normal  depth,  h  =  x0  cos  i  — —  = 

sin  (i  —  <p) 

20°  c  cos  (2  i  —  <p) 

10.8   ft.,   the   passive  thrust  =  active  thrust  = 


5°  sin  (i  -  <p) 

14  Ibs.  per  sq.  ft.,  ft'  =  -  (i  -  <p)  =  -  5°,  y  =  90  +  *  =  115°- 
•  cos  (2  i  —  <p) 

At  the  normal  depth,  h  =  — ; — — =  9.94  ft.,  the  passive  thrust  = 

sin  (i  —  <f>) 

c  cos  <p 

=1080  Ibs.  per  sq.  ft.  and  ft'  =  o,  y'  =  90°  +  <f>  =  110°.     For 

n  (i  —  <f>) 

=  5  ft.,  prove  by  constructing  the  circular  diagram  that  the  passive  thrust 
=  810  Ibs.  per  sq.  ft.,  ft'  =  11°,  y'  =  99°.  The  unit  passive  thrust  in- 
eases  from  the  surface  down  to  the  normal  depth  9.94  ft.  and  then  decreases 

)wn  to  h  =  10.8  ft. 

. 

There   is   a   passive   thrust  when   i  >45°  +  —  but  it  is  of 

nail  interest,  since  then,  there  is  no  active  thrust,  Art.  90, 
ence  it  will  simply  be  stated,  without  proof,  that  the  curves 
T',  A'S',  then  extend  only  to  the  depth,  x'  =  A  A',  when  ten- 
on is  excluded. 

In  connection  with  Fig.  61,  let  A'B  represent  a  mountain 

ope,  and  suppose  a  break  in  the  upper  part,  due  to  excessive 

lins,  say;  then  if  the  resulting  pressure  is  sufficient  to  overcome 

le  passive  resistance  of  the  part  below,  the  upper  part  will 

ide  along  a  surface  similar  to  A  'AS  and  the  lower  part  along 

surface  similar  to  S'Af.    If  the  earth  below  the  slide  is  of 

much  firmer  consistency  than  that  above,  the  moving  earth 

ill  flow  downward  over  A'  and  along  the  free  surface. 


182     EARTH   ENDOWED   WITH   BOTH   COHESION   AND   FRICTION 


96.  Active  Thrust.     Surface  Horizontal.     See  Fig.  62,  i  =  o. 
In  this  case,  r  =  OR  =  OJ  =  a  =  wx  and  the  conjugate  thrust 


C  col 


FIG.  62 

r'  =  01  =  b.  On  dropping  the  perpendiculars  IE  and  JF 
upon  the  tangent  KS,  we  find  IE  =  (b  +  c  cot  <p)  sin  <p,  JF  - 
(a  +  c  cot  <p)  sin  <p 

.'.  2  PS  =  (a  -\-  c  cot  (p  -\-  b  -\-  c  cot  <p)  sin  <p  =  a  —  b. 

i  —  sin  (p  sin  (go0  —  <p) 

.'.  b  =  a 2  c 


i  +  sin  (p 
a  tan2  (45°  -  -)  -  2 


cos  (go0  —  (p) 
2  sin  (45° J  ow  (45°-  ' 


a  CM*  (45°    ' 

.'.  b  =  a  tan2  (45° J  —  2  c  tan  ^45° J 

.'.  6  =  tow  (45°  -  -)  [ra  tow  (45°  -  -)  -  2 


2/i-  \  2< 

Here  a;  is  the  vertical  depth,  below  the  free  surface,  at  which 
the  horizontal  unit  thrust  =  b. 

Writing  this  equation  in  the  form, 

b  =  w  tan2  (45°  —  - )  \  x tan  (45°  + 

V  ~         2/  L         w          \ 

2  C  f  *P\ 

and  putting  x'  =  —  tan  (45°  +  -),  we  have 

W  V  2X 

b  =  wtow2  (45°  --)   [x-xr]. 
\  »/ 


96] 


ACTIVE   THRUST.      SURFACE    HORIZONTAL 


183 


b  =  o  when  x  =  x' ',  which  agrees  with  the  result  of  Art.  92. 

The  equation  is  not  valid  for  x  <  x',  for  an  unlimited 

mass  of  earth. 

The  planes  of  rupture  75  and  IT  have  their  true  directions, 
since  //  is  now  horizontal.  Therefore  the  angle  with  the  verti- 
cal IN,  Fig.  62,  made  by  either  plane  =  NIS  =  K  IPS  =  K 
(90°  —  <p) ;  so  that  a  plane  of  rupture  bisects  the  angle  between 
the  vertical  and  the  natural  slope.  Since  this  result  is  true  for 
any  x  <  x' ,  it  follows  that  the  plane  of  rupture  maintains  the 


FIG.  63 

same  inclination  DAI,   Fig.   63,  with  the  vertical,  from  the 
level  ID,  where  x  =  x',  down. 

In  Fig.  63,  where  BC  is  the  level  free  surface,  on  putting 


b  =  wytan2  (45°  -— 
\  v         ? 


d) 


which  is  the  formula  for  the  horizontal  unit  thrust  of  non- 
coherent earth,  with  ID  for  a  free  surface. 

On  laying  off  the  value  of  b  at  A  =  FA  and  drawing  a  straight 
line  from  /  to  F,  it  is  seen  that  the  total  horizontal  thrust  E  on 
the  vertical  plane  AB  is  represented  by  the  area  of  the  triangle 

IAF  =  —  yb 

2 

.'.  E  =  —  wy*  Ian2  (45°  -—)...      (2) 

2  \  2  ' 

Also,  since  it  is  a  uniformly  varying  stress,  the  resultant 


184    EARTH  ENDOWED   WITH  BOTH  COHESION  AND  FRICTION 


acts  at  y$  y  =  */£  A I  above  A .  To  compute  the  numerical  value 
of  E  for  a  given  x,  we  first  compute, 

2  c         /     o        ^  \ 

x  =  — /aw  (45    H ), 

w;          v  2  / 

then  y  =  #  —  #',  and  finally  £  for  the  given  values  oi  c,  w 
and  <p. 

For  an  unlimited  mass  of  earth,  subjected  to  no  force  but 
its  own  weight,  it  will  now  be  shown  that  there  is  no  horizontal 
stress  on  a  vertical  plane,  in  the  layer  IDCB  of  Fig.  63  of  depth  x'. 

In  Fig.  64,  let  DE  be  the  free  surface  and  at  the  depth  x  = 
DB,  conceive  a  wedge  of  rupture  ABC,  the  angle  BAC  being 

(p 

45° ,  as  proved  above.     If  we  conceive  the  wedge  to  be 

2 

without  weight,  the  pressures  on  the  three  faces  will  correspond 
to  those  at  the  depth  x. 

Let  AB  =  i,  BC  =  d  and  AC  =  /;  also  let  n  =  unit  normal 
reaction  of  earth  on  plane  AC]  thus  the  total  normal  reaction  is 
nl,  the  frictional  resistance  is  nlf  =  nl 
tan  <p  and  the  cohesion  d,  the  last  two 
forces  acting  up  along  AC,  opposite  to 
the  impending  motion.  The  wedge  is 
thus  in  equilibrium  under  the  forces, 
b  X  i,  wxd,  nl  and  d  +  nlf.  If  we  lay 
off  NiOLi  =  <p,  then  if  nl  =  ON1}  LiNi 
=  nlf  or  if  nl  =  ON2,  then  L2N2  =  nlf, 
etc.  Prolong  N\L\  to  Q\,  making  Qi 
LI  =  d  and  draw  QiQs  parallel  to  OL\. 
Now  if  wxd  =  OPij  then  the  hori- 
zontal PiQi  represents  *>,  since  we  have 
a  closed  polygon  OPiQiNiO  whose  sides 
represent  the  forces  wxd,  b,  d  +  nlf  and 
nl,  in  order.  Here  b  =  PiQi,  acting  to  the  right,  is  compres- 
sive.  When  wxd  =  OP2,  the  force  polygon  is,  OPZN2O  and 
b  =  o.  This  corresponds  to  the  depth  x'.  For  a  less  depth,  so 
that  wxd  =  OP 3,  the  force  polygon  is  OP3N30,  and  since  now 
=  nl,  it  will  be  observed  that  P3N3  <  cl  +  nlf  =  d  + 


36]  ACTIVE    THRUST.       SURFACE    HORIZONTAL  185 

' 

LZN3.  Hence  when  x  <  x' ,  the  full  friction  and  cohesion  is 
not  exerted  on  the  plane  AC  and  its  resultant  reaction  P£> 
(the  resultant  of  PzN*  and  N£>)  is  vertical  and  exactly 
balances  the  weight  of  the  prism  BE  of  earth  resting  on  BC. 

There  is  thus  no  necessity  for  tensile  forces  and  none  are 
exerted  in  the  layer  IDCB  of  Fig.  63.  There  is  then  no  stress 
on  the  vertical  plane  DE,  Fig.  63,  since  b  =  o  for  any  point  of  it, 
and  each  vertical  prism  resting  on  DC  is  sustained  by  a  vertical 
reaction  of  the  plane  DC;  in  fact,  the  part  I  DEB  can  be  removed 
and  the  prism  DEC  will  be  entirely  sustained  by  the  vertical 
reaction  of  DC.  It  has  been  seen  too  that  only  a  portion  of  the 
full  cohesion  and  friction  possible  is  required  for  equilibrium, 
and  this  portion  becomes  proportionately  less  as  x  (or  OP3, 
Fig.  64)  diminishes. 

The  case  is  different  if  the  earth  gives  way  to  the  left  of  the 
vertical  plane  AB,  Fig.  63,  as  from  the  failure  of  a  retaining 
wall  or  from  digging  a  vertical  trench  at  a  greater  depth  than 
x1 '.  In  this  case,  if  the  earth  is  capable  of  exerting  tension,  the 
polygon  of  forces  in  Fig.  64,  when  wxd  =  OP3,  will  be,  OP3  Q3 
MO,  if  the  full  cohesion  and  friction  QZM  on  A C  is  exerted.  Since 
PdQs  is  directed  to  the  left,  b  =  P3()3  is  tensile.  If  the  reaction 
b  is  not  horizontal,  but  directed  upward,  as  is  probable,  the 
polygon  will  be  correspondingly  changed. 

Recurring  now  to  the  case  of  the  unlimited  mass  of  earth, 

where  there  is  no  stress  on  ED,  Fig.  63,  the  prism  EDC  being 

self  supporting,  it  is  seen  that  the  only  resistance  to  sliding  of 

the  prism  A  DEB,  down  the  plane  AD,  is  the  total  cohesion 

and  friction  exerted  on  AD  alone.      Stating  the  conditions  of 

equilibrium  for  the  prism  A  DEB,  subjected  to  its  own  weight, 

:.'the  total  reaction  on  AD,  and  the  thrust  on  AB,  acting  from 

ijleft  to  right,  it  can  be  proved  that  the  plane  of  rupture  AD 

makes  the  angle  I A  D  =  (45° J  with  the  vertical  and  that 

the  thrust  is  exactly  the  value  of  E  given  above.  The  proof, 
being  long,  is  omitted. 

If,  however,  the  earth  gives  sufficiently  to  the  left  of  AB, 


186     EARTH  ENDOWED  WITH  BOTH  COHESION  AND  FRICTION 

then  (i)  it  may  be  supposed  that  tensile  forces  will  be  exerted 
along  ED  that  will  drag  the  whole  or  a  part  of  the  prism  DEC, 
down  the  plane  DC  and  either  the  whole  or  a  part  of  the  co- 
hesion and  friction  that  can  be  exerted  along  DC  will  be  called 
into  play. 

If  the  full  amount  of  friction  and  cohesion  possible  is  exerted 
along  DC,  then  the  wedge  of  rupture  is  no  longer  A  DEB,  but 
ACB.  Stating  the  conditions  of  equilibrium  for  ACB,  it  can 
be  proved  that  the  equilibrating  horizontal  thrust  is, 

E'  =  —  x2  tan*  (45°  -  — }  -  2  ex  tan  (45°  -  — ) 

2  V"  2  '  ^  2  ' 

The  state  of  stress  throughout  the  mass  is  now  complex,  in- 
volving tensile,  compressive,  and  shearing  forces. 
When  Er  —  o,  we  derive, 

40 t      (     o   ,     ^\ 
x  =  —  tan  { 45    H 1 

W  \  2  / 

which  is  double  the  height  xr  found  above.  This  formula  will 
be  proved  independently  in  the  next  article. 

(2)  It  may  be  supposed  that  the  tensile  strength  of  the 
earth  along  DE,  Fig.  63,  is  not  sufficient  to  drag  EDC  down, 
but  that  the  impending  motion  of  A  DEB  is  resisted  not  only 
by  the  friction  and  cohesion  along  AD}  but  also  by  the  cohesive 
resistance  acting  upward  along  DE. 

(3)  After  heavy  rains,   the  earth  on  drying  out,  causing 
contraction  near  the  surface,  may  develop  vertical  cracks,  so 
that  not  even  cohesive  resistances  can  be  assured  along  DE. 

It  is,  perhaps,  a  common  fact  of  observation,  when  slides 
occur  in  trenches  or  open  cuts,  in  ordinary  clayey  earth, 
that  the  surface  of  rupture  is  nearly  vertical  for  several  feet 
down  from  the  top,  and  below  this  it  is  much  less  inclined  to 
the  horizontal.  The  lower  part  of  the  break  is  often  curved 
and  concave  upwards.* 

*  In  the  discussion  of  Mr.  J.  C.  Meem's  paper  on  "The  Bracing  of  Trenches 
and  Tunnels,"  Trans.  Am.  Soc.  C.E.,  vol.  LX,  pp.  I — 100,  a  great  many  facts  of 
observation,  vital  to  the  constructor,  are  recorded.  In  this  connection,  see 
especially  Mr.  E.  G.  Haines's  remarks  on  surfaces  of  rupture. 


8,  97]  TRENCH  WITH  SLOPING  FACE  187 

Such  facts  lend  color  to  the  belief  that  either  assumptions 
2)  or  (3)  are  more  probable  than  (i).  To  be  on  the  side  of 
afety,  (3)  is  suggested,  so  that  it  is  well  not  to  count  on  a  greater 
eight  than  x'  for  the  unbraced  trench,  particularly  for  a  period 
f  years. 

An  important  case,  often  met  with  in  practice,  is  that  of 
he  retaining  wall,  backed  by  earth  level  with  the  top  of  the 
rail,  the  earth  being  loaded  uniformly.  Conceive  the  sur- 
harge  to  consist  of  the  same  weight  of  earth  whose  specific 
weight  is  that  of  the  filling,  the  height  of  such  a  surcharge 
•eing  x0.  Let  the  inner  face  of  the  wall  pass  through  A, 
"ig.  63,  and  lie  to  the  left  of  the  vertical  through  A  and  sup- 
•ose  it  required  to  find  the  earth  thrust  on  the  vertical 
•lane  from  A  to  the  surface  of  the  filling,  the  thrust  acting 
.orizontally. 

Let  BC  represent  the  top  of  the  surcharge,  so  that  the  hori- 
ontal  surface  of  the  filling  is  at  a  distance  x0  below  BC.  Then 
I  x0  <.  x' ,  or  if  the  surface  of  the  filling  is  at  ID  or  above  it,  then 
he  thrust  E  required  is  found  from  (2)  above,  since  AD,  where 
he  resistance  to  sliding  is  exerted,  lies  wholly  in  the  filling. 
This  is  not  true  when  x0  ^  xf,  for  then  a  part  of  AD  lies  in  the 
urcharge  which  offers  no  resistance  to  sliding.  In  this  case, 
ompute  by  (i),  the  unit  pressure  at  depth  x0  or  at  the  surface 

the  filling;  also  compute  by  (i)  the  unit  pressure  for  x  =  AB 
»r  from  A  to  the  surface  of  the  surcharge.  The  pressure  in- 
reases  uniformly  from  x  =  x0  to  x  =  AB  and  the  resultant 
hrust  and  its  point  of  application  can  be 
ound  as  in  the  corresponding  problem  re- 
ating  to  Fig.  52,  Art.  88. 

97.  Trench  with  Sloping  Face.  In  Fig. 
»5,  AB  represents  the  face  of  the  trench 
»f  vertical  height  h  and  inclined  at  an 
ingle  0  to  the  horizontal.  In  the  usual 
.heory,  any  trial  plane  of  rupture  AC,  in-  pIG 

:lined  at  the  angle  a.  to  AB,  is  supposed 
;o  exert  frictional  and  cohesive  resistances  throughout  its  whole 
extent  to  impending  motion  downward  of  the  wedge  ABC. 


188     EARTH  ENDOWED   WITH   BOTH   COHESION   AND   FRICTION 

W 

The  weight  W  of  ABC  =  —  AB.  AC  sin  a 

2 

W         H  AT      ' 

= ; —  .  AC  sin  a. 

2   sin  p 

The  components  of  W,  parallel  and  perpendicular  to  AC 
are,  W  sin  (ft  —  a),  W  cos  (p  —  a),  respectively  and  the  fric- 
tional  resistance  along  AC  will  be,  W  cos  (p  —  a)  tan  (p. 

The  cohesive  resistance  =  c'  X  AC,  if  we  call  c'  the  cohesion 
per  unit  area  on  the  plane  AC.  For  the  true  plane  of  rupture, 
the  cohesion  is  c,  the  maximum,  or  coefficient  of  cohesion.  For 
any  other  plane  as  AC,  c'  <  c  and  c'  varies  with  a. 

Balancing  components  parallel  to  AC,  for  equilibrium, 
W  [sin  (p  -  a)  -  cos  (p  -  a)  tan  «*]  =  .c'  X   AC. 

sin  (p  —  a  —  <p) 

The  [  ]  reduces  to . 

cos  <p 

Substituting,  replacing  W  by  its  value  and  dividing  by  AC, 

w  h  sin  a  sin  (p  —  a  —  <p)        f 

2    sin  P  cos  (f> 

dc' 

This  is  a  maximum  when  —  =  o. 

da 

.".  cos  a  sin  (p  —  a  —  (p)  —  sin  a  cos  (p  —  a  —  <p)  =  o 

.'.  tan  a  =  tan  (p  —  a  —  <p)  .'.  a  =  p  —  a  —  <p. 
P  —  <p 

.  .     OL   —    ~~ 

2 

or  AC  bisects  the  angle  (p  —  <p)  =  BAS. 

On  substituting  this  value  of  a  in  the  value  of  c\  which  now 
becomes  c.  we  find 

(3  —  <P 
wh  sin2  -       -  =  2  c  sin  p  cos  <p. 

2 

To  criticize  this  well-known  formula  for  computing  the 
maximum  height  h  for  a  stable  slope,  let  the  vertical  DE  = 

—  tan  [45°  +  — ) ;  then  before  excavating  the  trench,  there  is 
w        v  "        a' 

no  stress  on  DE,  Art.  92.    After  the  earth  is  removed  to  the 
left  of  AB,  the  mass  A  DEB  would  slide  down  AD,  unless  the 


]  TRENCH   WITH   SLOPING   FACE  189 

irth  along  DE  could  exert  sufficient  tension  to  drag  down  the 
hole  of  the  wedge  EDC,  which  is  improbable,  since  observation 
lows  that  the  upper  part  of  a  slide  is  nearly  vertical.  Hence  the 
irf ace  of  rupture  is  not  A  DC,  but  say  A  DE  or  a  curved  surface 
ing  perhaps  near  it.  Hence  the  true  height  for  an  unsupported 
•ench  or  open  cut  is  less  than  is  given  by  the  above  formula. 
To  deduce  a  formula,  where  the  trial  prism  of  rupture  is  of 
le  type  A  DEB,  let  i  =  inclination  of  BC  to  the  horizontal  and 
enote  DE  by  x'\  also  draw  from  D  a  parallel  to  BC  intersecting 
(B  at  /.  Replace  the  area  I  DEB  by  the  parallelogram 
sides  ||  ID  and  ED)  whose  area  is  ID  X  DE  cos  i,  which 
>  on  the  side  of  safety.  The  weight  of  the  prism  A  DEB  is  thus, 

w.ID[y2  AI  sin  (ft  -  i)  +  x'  cos  i]  = 

. 

w  .  AD      Sma    ^  [y2  AI  sin  (ft  -  i)  +  xf  cos  i]. 

sin  (ft  —  i) 

This  expression  replaces  W  in  the  first  equation  of  equilibrium 
,bove  and  c'  .AD,  replaces  c' '.AC.  After  dividing  the  resulting 
quation  by  AD,  we  have, 

it; .  sin  a  sin  (ft  —  a  —  <p)  r  . 

: — VA  AI  sin  (ft—  i)  +  x'  cos  i]  =  c'. 

sin  (ft  —  i)  cos  0 

On  differentiating  with  respect  to  a,  we  find  as  before,  for  a 
naximum  cf,  a  =  ^  (ft  —  <p) ;  whence  substituting  and  solving 
or  AI, 

2  C        COS  (p  COS  I 

AI  = —  2  x 


w    .     ft  —  <p  sin  (ft  —  i) 

sin2  ~ 
2 

Also,  since  h  —  AI  sin  ft  +  IB  sin  ft  = 

,  sin  ft  cos  i 

AI  sin  ft  +  x'  -— -, 

sin  (ft  —  i) 

we  reach  the  final  formula, 

2  c  cos  <p  sin  0         ,  sin  ft  cos  i 
h  = —  x 


w 
sin 

2  c 


ft  —  <p  sin  (ft  —  i)' 


1C  (  (p  \ 

ere,  x  =  —  tan  (  45    H ). 

1H  V  V    ' 


190    EARTH  ENDOWED   WITH  BOTH  COHESION  AND  FRICTION 

It  will  be  observed  that  this  value  of  h  only  differs  from  that 
first  found  in  the  subtractive  term.  As  an  application  let  i  =  o 
c  =  300  lb./ft.2,  ze;  =  112  lb./ft.3,  <p  =  8°,  ft  =  33°  41',  whence 
by  the  formula  h  =  53.6  ft.  This  only  differs  xf  =  6.16  ft. 
from  the  result  of  the  first  formula. 

When  c  is  taken  as  400  lb./ft.2,  the  other  quantities  remain- 
ing the  same,  we  find  h  =  71.4  ft.  In  the  reservoir  embank- 
ment of  Charmes,*  which  failed,  the 
experimental  determinations  were  <p  = 
8°,  ft  =  3304i',  c  =  400  lb./ft2  (395  to 
448),  w=-ii2  lb./ft.3,  h  =  56  ft.  or  less 
than  71.4  as  found  by  the  formula. 
Apparently  the  formula  still  gives  an 
excess  height,  particularly  so  since  the 
lower  20  ft.  of  the  embankment  was 
covered  with  water,  whose  hydrostatic 

pressure  added  to  the  stability.  The  surfaces  of  rupture  were 
somewhat  of  the  shape  shown  in  Fig.  66,  by  the  dotted  line. 
The  ruptures  were  attributed  to  the  lowering  of  the  water  level. 
If  in  the  last  formula,  we  make  ft  =  90°,  we  have  for  the 
maximum  height,  h0  at  which  a  vertical  trench  will  stand  un- 
supported, when  i  =  o, 

2  c  sin  (90°  —  <p)          , 

n0  = x   = 

w  /  &  \ 

*t  (45? -4-] 

sin  (45°  -  -*-)  cos  (45°  -  ^ 

4C  \  2  '  V  2 

<  * 

/.  h0  =  ^  tan  (45°  *.-f.)  _y  -•  if  to  (45°  +  -*-). 

2t>\  jj  /  W\  2  ' 

As  before  remarked,   this  ignores  any  possible  tension  in 
the  earth  along  DE,  Fig.  65.     If  sufficient  tension  exists  to  drag 


*  Resal,  Poussee  des  Terres,  II.,  pp.  327-340. 


:|  TRENCH   WITH   SLOPING  FACE  191 

:>wn  the  wedge  EDC,  then  the  last  value  of  h0  given  will  be 
oubled,  since  the  term  x'  is  then  omitted. 

A  further  restriction  must  be  placed  upon  the  angle  i  that 
t.e  surface  BC  makes  with  the  horizontal,  namely,  that  gen- 
tally,  i  ^  <f>\  for  if  i  >  <f>,  and  if  a  line  be  drawn  parallel  to  BC 
)    I  the  critical  depth  xoj  Art.  91,  and  this  line  intersects  AB, 
i    jtere  will  be  sliding  along  the  corresponding  surface  at  the 
I    epth  x0  below  BC.     Generally  (or  when    ft  is  not  near  90°) 
.,  <  h,  so  that  i  should  be  i  <p  for  stability. 

Considering  the  face  AB,  Fig.  66,  making  the  angle  ft  with 
le  horizontal,  the  critical  depth  normal  to  AB,  is  Art.  91, 

c         cos  ( 
ED  =  x0  cos  ft  =  — 


w  sin  (ft  —  <p) 

This  is  strictly  true  only  when  AB  is  unlimited  and  is  only 
pproximately  true  for  usual  heights  of  trench.  Sliding  is  then 
nminent  in  the  vicinity  of  point  D,  Fig.  66,  the  surface  of 
ipture  passing  through  A  and  .having  about  the  shape  A  DC, 
eing  curved  and  concave  upward.  Resal,*  who  suggests  this 
urve  ADC,  further  states  that  it  should  make  an  angle  of 

(p 
5° with  AB  at  A,  since  the  plane  of  rupture,  for  passive 

esistance  of  the  earth  at  A,  makes  this  angle  with  the  surface 
egarded  as  AB.    Thus  in  Fig.  61,  the  curve  A'T'  makes  the 

ngle  (45°  -  — )  with  AfB  at  A'j 

The  bottom  of  the  trench  to  the  left  of  A  can  make  an 
.ngle  with  the  horizontal  i\,  provided  ii  .<  <?,  for  in  that  case 

*  Poussee  des  Terres,  II.,  p.  95. 

fin  applying  the  formula  for  ED  to  the  Charmes  embankment,  which  was 
omposed  of  heterogeneous  materials  and  constructed  after  the  most  ap- 
>roved  manner,  it  was  thought  at  D,  the  earth  being  firmer  and  less 
;aturated  than  near  the  surface,  that  <p  =  9°  40'  and  possibly  c  =  800  Ib.  ft.2, 
vould  represent  more  nearly  the  constants.  On  substituting  in  the  formula, 
ve  find  ED  =  17.3  ft.  In  the  actual  slide  ED,  as  measured,  was  24.6  ft.,  the 
turface  of  the  slide  taking  the  general  shape  of  the  dotted  curve  in  Fig.  66  and 
imerging  at  the  upper  surface  nearly  at  right  angles  to  it. 


192     EARTH  ENDOWED   WITH  BOTH   COHESION  AND   FRICTION 


the  passive  resistance  of  the  earth  to  the  left  of  a  vertical  plane 
AF  is  sufficient  to  overcome  the  active  thrust,  on  this  plane, 
of  the  earth  to  the  right  of  FAB;  but  this  is  not  true  when 
ii  >  <p,  since  at  the  critical  depth  x0  =  AF  (putting  ii  for  i  in 
formula)  the  equilibrium  is  unstable  and  for  a  greater  depth 
there  is  no  equilibrium,  so  that  heaving  of  the  earth  in  the 
bottom  of  the  trench  would  occur. 

It  will  be  observed  from  the  formula  for  ED,  Fig.  66,  that 
when  j8  =  90°,  ED  =  c/w. 

It  may  readily  be  surmised  from  what  precedes  that  the 
theory  of  stable  slopes  is  not  on  a  very  satisfactory  basis,  but 
its  indications  may  be  of  service  to  the  engineer  when  coupled 
with  his  practical. judgment. 

98.  Earth  Surface  Horizontal.  Passive  Thrust.  Referring 
to  Art.  96  and  Fig.  62,  it  is  seen  that  when  r  =  01  =  b  =  wx, 
then  the  passive  horizontal  thrust,  /  =  OJ  =  a;  hence  on 
solving  an  equation  in  Art.  96,  for  a,  we  derive, 


a  =  b  tan2  (45°  +  ~)  +  2  c  tan  (45°  +  — 


.'.  putting  x"  =  —cot  (45°  +  —  ), 
to        \  2  / 

a  =  w  (x  +  x")  tan2  (45°  +  —V* 

99.  Foundations.     In  Fig.  67  is  shown  a  wall  backed  b] 
earth.    Let  W  =  vertical  component  of  the  resultant  on  the 
earth  foundation  BC  in  pounds  and  A  =  area  base  EC  in  square 
feet,  both  for  i  foot  length  of  wall.     If  the  resultant  cuts 
base  at  its  center,  the  vertical  unit  pressures  p  and  p'  at 
toe  and  heel  of  the  wall,  each  equal  W/A\  o^nerwise  they 

*  If  earth  is  placed  in  front  of  a  stable  wall,  only  its  active  thrust  is  exerted 
against  the  wall.  If,  however,  the  wall  slides,  even  a  very  small  distance,  the 
full  passive  resistance  is  brought  into  play,  which,  for  earth  level  at  top,  can 
be  computed  by  aid  of  the  above  formula,  which  gives  the  unit  pressure  at  the 
surface  for  x  =  o  or  generally,  for  any  depth  x,  the  pressures  varying  as  the 
ordinates  of  a  trapezoid  from  the  surface  down.  The  resultant  horizontal 
pressure  thus  acts  through  the  center  of  gravity  of  the  trapezoid. 


99] 


FOUNDATIONS 


193 


be  computed  from  a  formula  given  in  Art.  15,  so  that  it  will  be 

assumed  that  both  are  known  for  any  position  of  the  resultant. 

Let  wi  and  w*  denote  the  weight  per  cu.  ft.  of  the  earth 

in  front  of  the  wall  and  of  the  filling  back  of  the  wall  respectively. 


b  i  y 

FIG.  67 


The  coefficients  c  and  /  =  tan  <p,  refer  to  the  earth  of  the  wedges 
of  rupture  just  below  BC. 

The  active  thrust  b  at  the  toe  of  the  wall,  caused  by  the 
pressure  p,  tending  to  cause  the  little  wedge  of  rupture  to  slide^ 
down,  is,  by  a  formula  of  Art.  96, 

b  =  p  tan*  (45°  -  ")  ~  2  c  tan  (45°  -    — ). 

If  x  denote  the  depth  from  the  surface  of  earth  in  front  of 
the  wall  to  B,  then  if  b  is  sufficiently  large  to  develop  the  full 
passive  resistance  of  the  earth  just  to  the  left  of  jB,  Art.  98, 

:wi  x  tan2  (45°  +  — )  +  2  c  tan  (45°  H ), 
\                      2  '                               ^                      2  ' 
to  cause  impending  motion  of  the  corresponding  wedge  of 
ture  up  the  plane  of  rupture,  so  that  the  earth  is  about  to 
ve  or  rise,  then  the  active  thrust  is  equal  to  the  passive 
thrust.     Placing  them  equal  and  solving  for  p,  we  have,* 


tan*  (  45°  +  -  -)  +  2  c  tan*  (  45°  +  '  ~ 
(  45°  +  -- 


2  c  tan 


(0 


*  Reducing  by  relation,  tan  (45°  -  —  )  =  cot  (45°  +  —  J  = 

2/ 


194     EARTH  ENDOWED   WITH   BOTH   COHESION  AND   FRICTION 

The  unit  toe  pressure  must  not  exceed  this  value  of  />,  or 
heaving  of  the  earth  in  front  of  the  wall  will  occur. 

For  a  given  toe  pressure  p  the  equation  may  be  solved  for 
x,  the  least  height  the  earth  should  extend  above  B  to  prevent 
heaving. 

At  the  heel  of  the  wall  C,  where  the  height  of  filling  is  h,  the 
active  horizontal  unit  thrust  due  to  the  filling  is, 

w2  h  tan2  (  45° J  —  2  c  tan  (  45° J  . 

This  should  not  exceed  the  passive  unit  resistance  to  sliding 
up  the  plane  of  rupture  shown  in  the  figure  just  below  C, 

p'  tan*  (  45°  + —)  +  2  c  tan  (  45°  +  — ) , 

or  the  earth  will  heave  the  wall. 

Equating  the  active  and  passive  thrusts  and  solving  for  p', 

p'  =  w^h  tan*  (  45°  —  — j  —  2  c  tan3  (  45° j  — 

2  c  tan  (  45°  -  — ) (2) 

To  prevent  heaving  or  tilting  of  the  wall,  p'  should  not  be  less 
than  the  value  given  by  (2). 

In  deriving  (i)  and  (2),  the  horizontal  component  H  of  the 
resultant  thrust  on  BC  has  been  neglected,  though  it  un- 
doubtedly exercises  an  influence  on  the  pressures  both  at  B 
and  C. 

If  the  actual  pressure  at  the  heel  of  a  retaining  wall  is  less 
than  the  value  given  by  (2),  the  wall  will  tilt  over  somewhat; 
likewise  if  the  pressure  at  the  toe  is  greater  ihan  the  value 
given  by  (i),  the  earth  in  front  of  the  wall  will  heave  and  the 
toe  may  sink  slightly.  If  the  increased  leaning  of  the  wall 
from  one  or  both  causes  is  not  too  great,  its  decreased  stability 
may  be  so  small  that  the  wall  remains  perfectly  stable. 

In  fact,  heavy  weights,  as  machinery,  road  rollers,  etc.,  are 
constantly  being  placed  on  the  surface  of  the  ground  without 
mishap.  However,  the  soil  is  compressed  somewhat  and  the 


99,  100]  SCREW   PILE  195 

earth  may  heave  a  little  on  the  sides.  Similarly,  the  weight 
of  a  string  of  locomotives,  on  a  narrow  embankment,  is  safely 
transmitted  to  its  base.  Imagine  a  wall  to  replace  the  loco- 
motives. The  earth  is  likely  to  settle  more  than  where  earth  is 
filled  in  outside  the  embankment,  either  level  with  its  top  or 
rising  above  it,  as  in  Fig.  67.  In  the  latter  case,  when  x  is 
greater  than  the  value  given  by  (i),  for  a  given  p,  there  can  be 
no  sinking  at  the  toe,  from  the  little  wedge  of  rupture  there 
sliding  down. 

It  is  true  that  most  of  the  weight  of  the  wall  is  sustained  by 
the  earth  occupying  the  space  of  the  original  embankment, 
still  it  is  desirable,  for  a  stable  wall,  to  have  no  sinking  of  the 
kind  supposed  and  consequently,  the  formulas  above  may  prove 
useful  in  ascertaining  permissible  soil  pressures  p  and  pf  when 
the  constants  wi,  w2)  C  and  <p,  have  been  determined  by  ex- 
periment.* 

EXAMPLE.  Referring  to  Fig.  67,  let  w\  =  w2  =  ioo  lb./ft.3,  the  specific 
weight  of  earth  before  and  behind  the  wall.  For  the  earth  of  the  foundation, 
just  below  level  BC,  let  <p  =  15°,  c  =  200  lb./ft.2.  Also  let  *  =  3  ft.  h  = 
20  ft. 

Therefore  by  (i)  the  pressure  p,  at  the  toe,  must  not  exceed  2271  lb./ft.2, 
or  the  earth  in  front  of  the  wall  will  heave,  and  by  (2)  the  pressure  p'  at  the 
heel  must  not  be  less  than  205  lb./ft.2  or  the  earth  will  heave  the  wall.  When 
(2)  gives  p'  negative,  it  indicates  that  the  active  thrust  is  less  than  the  pas- 
sive, so  that  no  heaving  of  the  wall  can  occur  even  when  there  is  no  pressure 
of  the  masonry  at  the  heel. 

ioo.  Screw  Pile.  The  greatest  unit  pressure  p,  acting 
vertically,  that  can  be  sustained  by  the  earth  at  the  base  of  a 
screw  pile,  without  heaving  of  the  earth  or  settlement  of  the 
pile,  is  given  by  (i)  of  Art.  99,  x  being  the  depth  from  the  surface 
down  to  the  screw. 

Thus  for  a  moist,  clayey  earth,  suppose  <p  —  15°,  Wi  =  ioo 
lb./ft.3,  c  =  400  lb./ft.2,  then  from  (i),  Art.  99,  we  find, 
p  =  288  x  +  2768  lb./ft.2 

*  See  paper  by  Mr.  A.  L.  Bell  on  "The  Lateral  Pressure  and  Resistance  of 
Clay  and  the  Supporting  Power  of  Clay  Foundations,"  Min.  Proc.  Inst.  C.  E., 
vol.  CXCIX,  1914-1915,  Part  i,  where  similar  formulas  to  the  above  were 
first  given  and  applied  in  the  manner  indicated. 


196   EARTH   ENDOWED   WITH   BOTH   COHESION   AND   FRICTION 

If  A  =  area  horizontal  projection  of  the  screw  in  square  feet, 
the  load  the  pile  can  just  carry  is  p  A  pounds.  The  same  for- 
mula applies  in  finding  the  unit  pressure  that  can  be  sustained 
by  the  earth  at  the  foot  of  a  cylindrical  pile.  The  friction 
exerted  on  the  sides  of  the  pile  from  the  earth  pressure  cannot 
be  computed  with  any  accuracy,  possibly  from  a  polishing  of 
the  earth  next  the  pile  and  a  consequent  large  decrease  there 
in  the  coefficients  c  and  /  as  the  pile  is  driven  down  or  sinks. 
The  sustaining  power  due  to  the  friction  of  the  earth  on  the 
vertical  sides  of  the  pile  is  usually  estimated  empirically. 

101.  Thrust  Against  a  Retaining  Wall  in  the  Case  of  Co- 
herent Earth.  If  we  proceed  in  a  manner  analogous  to  the 
usual  Rankine  method,  to  find  the  thrust  against  the  battered 
wall  of  Fig.  67,  the  thrust  which  acts  on  a  vertical  plane 
through  the  heel  C  is  first  ascertained  and  this  thrust  is  then  com- 
bined with  the  weight  of  earth  lying  between  the  vertical  plane 
and  the  wall,  to  get  the  thrust  on  the  latter.  When  the  surface 
is  horizontal,  the  horizontal  thrust  on  the  vertical  plane  is  given 
by  the  value  of  E,  Art.  96,  and  it  acts  %  (h  —  x')  above  C,  since 
it  is  uniformly  increasing  from  the  depth  x'  (where  it  is  zero) 
to  the  heel. 

If  we  conceive  a  plane  drawn  parallel  to  the  surface  and  at 

2  C  /  (f>\ 

a    deptk     x'  = —  tan  (45°  -\ — ),   below  it,  and  regard  this 

W  ^  2' 

plane  as  the  free  surface  of  non-coherent  earth  of  the  same  specific 
weight  and  angle  of  friction  as  the  given  earth  and  compute  the 
thrust  on  the  vertical  plane  for  such  earth,  the  resulting  formula 
is  found  to  be  identical  with  the  value  for  E  for  the  coherent 
earth  just  quoted. 

The  same  conclusion  holds  approximately  when  i  <  <p,  if 
the  surface  is  inclined  at  the  angle  i  to  the  horizontal.  Thus 
for  the  example  of  Art.  93,  footnote,  where  i  =  15°,  <p  =  30°, 
c  =  w  =  100,  x'  =  3.46  ft.,  at  the  vertical  depth  x  =  517.50  ft., 
the  unit  thrust  as  given  by  the  circular  diagram,  is  found  to  be, 
r'  =  19,200  lb./ft.2.  By  Rankine's  formula,  Art.  48,  for 
x  =  517.50  —  3.46  =  514.04,  the  thrust  is  the  same.  At  the 


101]  THRUST   AGAINST  A   RETAINING   WALL  197 

depth  31.05  ft.  by  circular  diagram,  r'  =  1020,  but  for  #'«= 
31.05  —  3.46  =  27.59  ft.,  the  Rankine  formula  gives  r'  =  1029 
lb./ft2 

As  we  have  seen,  when  i  is  not  zero,  the  unit  thrust  is  not 
exactly  uniformly  increasing,  but  it  can  be  assumed  as  such 
with  all  desirable  accuracy,  when  i  <  y,  so  that  the  resultant 
thrust  can  be  assumed  to  act  %  (x  —  #')  above  C,  Fig.  67,  where 
x  is  the  vertical  distance  from  C  to  the  inclined  surface  of  the 
earth. 

By  reference  to  Fig.  64,  referring  to  the  case  where  i  =  o, 
Z  BAG  constant  for  any '  -x  >  xf,  it  is  seen  that  P\Q\  varies 
lineally  with  PiP2  or  the  conjugate  thrust  varies  uniformly  as 
(x  —  x').  Similarly  it  might  be  inferred  when  i  <  <p,  since  the 
inclination  of  the  surface  of  rupture,  Fig.  60,  varies  but  slightly 
with  x}  that  r1 ',  the  conjugate  thrust  varies  nearly  uniformly 
with  (x  —  xf),  being  zero  at  x  =  x' .  This  is  easily  seen  from 
Fig.  64,  regarding  Z  BA C  as  constant,  on  drawing  P\Q\  parallel 
to  the  inclined  surface  to  represent  rf . 

However,  when  i  is  very  near  <?  and  especially  when  i  =  <?, 
r'  is  not  found  to  vary  lineally  (see  Ex.,  Art.  89),  the  surface 
of  rupture  being  considerably  curved.  The  resultant  thrust  acts 
a  little  below  the  third  point,  so  that  it  is  safe  to  regard  it  as 
acting  at  the  third  point,  but  the  unit  thrust  at  C,  Fig.  67,  must 
be  found  from  the  circular  diagram  (see  remark  in  Art.  89)  and 
not  by  the  approximate  method  outlined  above.  When  i  >  <p, 
the  curve  AS,  Fig.  61,  differs  still  more  from  a  straight  line  and 
the  linear  law  does  not  obtain.  The  total  thrust,  when  x  ^L  x0, 
may  be  computed  as  explained  in  Art.  94  and  its  position  found, 
either  by  taking  moments,  or  assuming,  for  abundant  security, 
that  it  acts  ^  (x  —  x')  above  the  base. 

It  will  be  recalled  in  connection  with  Fig.  61,  that  for  a 
depth  greater  than 

C  COS  (f> 

:  -T— 

w  cos  ^  s^n  (i  —  <p) 

slipping  would  occur  along  AS  if  the  earth  was  free  to  move. 
However,  if  a  wall,  of  height  greater  than  x0,  retains  the  earth, 


198    EARTH   ENDOWED   WITH   BOTH   COHESION   AND   FRICTION 

the  reaction  of  the  wall  introduces  forces  not  contemplated  in 
the  theory  of  the  indefinite  mass  of  earth  subjected  only  to 
its  own  weight  and  the  previous  solution  does  not  strictly  apply. 
It  seems  reasonable  to  suppose  that  heaving  of  the  earth 
would  occur;  but  ignoring  that,  an  approximate  estimate  of  the 
thrust  can  be  made  by  supposing  the  conjugate  active  thrust 
/  to  increase  uniformly  from  rf  =  o  at  x  —  x'  to  the  value 
given  by  the  circular  diagram  or  formula  at  x  —  x0  and  that 
this  rate  of  increase  continues  to  the  bottom  of  the  wall.  Thus 
for  the  example  of  Art.  94,  footnote,  where  i  =  25°,  <p  =  20°, 
c  =  200  lb./ft.2,  w  =  100  lb./ft.3,  x'  =  5.71  ft.,  x0  =  23.80  ft., 
r'  =  1990  lb./ft.2  at  x  =  x0,  rf  can  be  assumed  to  increase  uni- 
formly from  o  at  x  =  x'  to  1990  at  x  =  x0  and  to  reach  the 

value  — X  1990  =  4400    lb./ft.2    at    x  =  45.71    ft. 

23-80  -  5-71 

Compare  with  the  result  of  the  last  example  in  Art.  102. 

The  indefinite  slope  does  not  occur  in  practice  for  filling 
deposited  after  the  wall  is  built,  since  embankments  are  rarely 
over  150  ft.  in  height.  In  this  case,  at  first,  the  surface  slopes 
perhaps  nearly  at  the  angle  of  repose  for  non-coherent  earth. 
In  course  of  time,  the  cohesion  of  the  earth  may  be  very  much 
increased  and  the  earth  thrust  diminished;  but  suppose  the 
earth  at  some  time  saturated  with  water  which  may  still  further 
increase  the  cohesion,  though  it  may  diminish  <p,  the  angle  of 
friction;  then  the  thrust  against  the  wall  may  be  greater  than 
in  the  first  instance.  Thus  in  the  Charmes  reservoir  dam,  after 
more  or  less  saturation  of  the  earth  with  water,  <f>  was  found 
experimentally  to  be  only  8°,  c  =  400  lb./ft.2  Presumably, 
33°  41'  was  the  value  of  <f>  for  the  dry  pulverulent  earth  when 
first  deposited. 

Comparing  the  horizontal  thrusts  on  a  vertical  plane,  for 
earth  surface  horizontal,  for  case  (i),  where  c  =  o,  <f>  =  33°  41', 
w  =  100  (say),  and  case  (2),  where  c  =  400,  <p  =  8°,  w  =  112 
lb./ft.3,  we  find  the  thrusts  the  same  at  the  depth  x  =  19.6  ft.; 
but  at  the  depth  46.3  ft.,  the  resultant  thrust  for  the  earth  of 
case  (2)  is  double  that  for  case  (i).  However,  for  case  (2), 


101]  THRUST   AGAINST   A   RETAINING   WALL  199 

it  acts  12.7  ft.  above  the  base,  whereas  for  case  (i)  it  acts  15.4 
ft.  above  the  base. 

Such  facts  offer  perhaps  an  explanation  of  the  failure  of 
retaining  walls  that  are  not  drained  and  of  bridge  abutments, 
where  the  filling  is  saturated  at  high  water.  Such  filling,  for  a 
little  distance  back  of  the  abutment,  should  preferably  be  of 
gravel,  whose  angle  of  friction  is  not  much  affected  by  water. 
In  the  case  of  a  face  wall  at  the  foot  of  a  long  mountain  slope, 
thorough  drainage  is  essential. 

It  may  be  further  remarked  that  in  any  case  when  i  >  <p, 
the  dangerous  height,  Art.  97,  should  not  be  exceeded,  other- 
wise the  earth  above  the  foot  wall  may  rupture  and  the  approxi- 
mate prism  of  rupture  slide  over  the  wall. 

A  novel  device  for  retaining  the  earth  of  a  railway  embank- 
ment, originated  by  Mr.  Gustav  Lindenthal,  is  described  in 
Engineering  News  of  May  6,  1915,  p.  886.  The  greatest  height 
of  the  embankment  is  65  ft.,  width  57  ft.  6  in.,  and  it  is  retained 
by  two  thin  vertical  concrete  walls,  57.5  ft.  apart,  which  are 
connected  every  10  ft.,  vertically  and  horizontally,  by  tie  rods 
2*4  in-  diameter,  which  are  anchored  into  the  concrete  by  longi- 
tudinal and  vertical  8-in.  i6l/^-\b.  channels.  The  earth  was 
rammed  as  it  was  filled  in  and  •  was  thoroughly  drained  with 
vertical  wells  and  horizontal  drains.  The  earth  thrust  was 
estimated  somewhat  arbitrarily.  A  more  correct  way,  for  this 
very  coherent  earth,  would  be  to  put  specimens  of  the  earth 
that  had  been  rammed  between  the  plaques  and  determine 
c  and  <f>,  as  explained  in  Art.  3,  and  then  use  the  method  given 
at  the  end  of  Art.  96  for  a  surcharged  filling,  in  estimating  the 
earth  thrusts  at  different  depths. 

Remark.  In  using  the  above  methods  of  estimating  the 
stability  of  a  retaining  wall,  it  is  understood  that  the  coefficients 
/  and  c  are  to  be  found  by  experiments  on  the  particular  filling 
used.  Then  the  stability  of  the  wall  can  be  tested  in  the  usual 
manner.  As  to  whether  the  wall  should  be  designed  for  the 
coherent  earth  depends  entirely  upon  the  degree  of  permanence 
of  the  coefficients  /  and  c. 

It  is  known  that  clay  is  a  treacherous  material,  which  has 


200    EARTH  ENDOWED   WITH   BOTH   COHESION  AND  FRICTION 


not  permanent  characteristics,  so  that  its  use  as  a  filling  is  not 
to  be  recommended;  but  it  is  possible  that  earth  containing 
sand  and  clay  with  over  50%  of  sand,  well  rammed  as  it  is  filled 
in  and  thoroughly  drained  by  both  vertical  wells  and  horizontal 
drains,  may  show,  under  the  stress  of  all  kinds  of  weather,  so 
little  variation  in  the  coefficients  that  they  may  be  regarded  as 
permanent,  and  thus  the  wall  can  safely  be  designed  for  the 
particular  coherent  earth. 

In  the  design  of  the  temporary  bracing  of  trenches,  the 
theory  for  coherent  earth  may  be  used,  but  experience  shows 
that  in  time — especially  after  heavy  rainfalls — the  values  of  / 
and  c  are  materially  altered,  so  that  bracing  so  designed  cannot 
be  regarded  as  a  permanent  feature. 

For  earth — even  clay — not  exposed  to  the  atmosphere,  as  in 
the  foundations  of  walls,  the  permanency  of  the  coefficients 
can  be  more  confidently  relied  on. 

102.  Graphical  Method  of  Estimating  the  Total  Thrust  of 
Coherent  Earth  Against  a  Retaining  Wall.  The  following 
method  of  estimating  the  total  thrust  against  a  wall  is  more 
general  than  the  preceding,  in  that  it  includes  the  influence  of 
the  friction  of  the  earth  on  the  wall;  besides  it  is  applicable  to 
walls  leaning  toward  the  earth  and  to  any  kind  of  a  surcharged 
wall.  Although  the  construction  below  is  given  for  a  particular 
case,  the  method  is  general  and  is  easily  applied  to  any  case  of 
a  battered,  leaning,  or  surcharged  wall. 

In  Fig.  68,  let  AB  be  the  inner  face  of  the  wall,  retaining 
the  earth  to  its  right,  whose  surface  is  represented  by  the  in- 
definite line  Bbi.  The  line  AH  is  horizontal. 

Let  (p  =  angle  of  friction,  c  =  coefficient  of  cohesion  in  lb./ft.2 
and  w  =  weight  of  a  cubic  foot  of  earth  in  pourds. 

(1)  Compute,  x'  =  —  tan  (45°  +  -)  and  lay  it  off  verti- 

W  \  2' 

cally  downward  from  the  free  surface.  Through  the  lower 
extremity  of  this  line  draw  a  parallel  7*4  to  the  free  surface. 

(2)  Lay  off  a  convenient  length  Fi  and  equal  lengths  12  = 
23  =  34,  along  7*4,  erect  verticals  i&i,  ib2,  .  .  .,  and  compute 


102] 


GENERAL   GRAPHICAL   METHOD 


201 


the  areas  Ai  bi  B,  A 2  b2  B,  .  .  .  Thus,  if  p  is  the  length  of 
the  perpendicular  from  A  upon  7*4,  area  A  i  bi  B  =  %  p.  Fi  + 
y2  (Fi  +  Bbi)  q,  where  q  =  perpendicular  distance  between  7*4 
and  Bb±.  Also,  area  A2  bz  bi  i  A  —  "12  (X  P  +  <?)  =  area  A$ 
b3  b2  2  A  =  area  A^  Z>4  b$  3  A.  By  continued  addition,  the 


FIG.  68 

areas  required  are  ascertained,  and  on  multiplying  by  w  the 
weights  of  the  trial  prisms  of  rupture, 

Ai  ^  B,  A2  &2  B,  A$  bs  B,  A$  bt  B, 
are  found. 

On  a  convenient  vertical,  as  Ag,  lay  these  weights  off  to 
scale  from  g  :  ggi,  gg2,  gg3,  gg^  representing  the  successive  weights. 

(3)  Measure  the  lengths  Ai,  Ai,  A$,  A^  to  the  scale  of 
distance  in  feet  and  multiply  each  length  by  c.    From  gi,  draw 
giHi  \\Ai  and  =  Ai.c,  to  the  scale  of  loads,  to  represent  the 
cohesion,  acting  upward  along  Ai.    Similarly,  draw  £2^2  \\A2 
and  =  c.^2,  gzHzllA^  and  =  c.A$,  g^n^\\A^  and  =  c.A$, 
to  represent  the  cohesion  along  A2,  A 3  and  ^4,  in  turn. 

(4)  With  A  as  a  center  and  a  convenient  radius,  as  Ah, 
describe  the  arc  kh  H.    With  g  as  a  center  and  the  same  radius, 


202    EARTH   ENDOWED   WITH   BOTH   COHESION   AND   FRICTION 

describe  the  arc  ssi.  Lay  off  H  A  d  =  <p  and  mark  the  inter- 
sections, #1,  a2,  .  .  .,  of  Ai,  A2, .  .  .  (produced  where  necessary) 
with  the  arc  kd  H.  Then,  with  dividers,  lay  off  chord  ssi  = 
chord  da\,  ssz  =  daz,  .  .  .,  whence  the  Hnes  gsi,  gs%,  .  .  .  make 
angles  <p  with  the  normals  to  planes  Ai,  A2,  A^,  A^,  Art.  19. 

(5)  If  we  assume  the  angle  of  friction  of  earth  on  wall  <pf 
to  equal  <?,  lay  off  chord  df  =  chord  hk,  whence  fA  gives  the 
direction  of  the  earth  thrust  on  the  wall,  making  the  angle  <p 
with  the  normal  to  AB.  Next,  draw  through  n\,  n%,  .  .  . 
lines  parallel  to  Af  to  intersections  ci,  c2,  .  .  .,  with  gsi,  gs2,  .  .  ., 
respectively.  Pick  out,  with  dividers,  the  longest  of  the  lines 
nc,  which  is  found  to  be  nz  cz  in  this  figure.  This  line,  to  the 
scale  of  loads,  represents  the  earth  thrust  approximately,  in 
pounds,  and  A2  is  the  corresponding  plane  of  rupture.  For 
greater  accuracy,  other  trial  planes  of  rupture  lying  between 
A  i  and  A  3  should  be  investigated  as  before,  to  find  the  maxi- 
mum nc,  otherwise  the  thrust  may  be  slightly  underestimated. 

Proof:  It  will  be  observed  that  if  we  regard  A2  as  the 
plane  of  rupture,  the  prism  of  rupture  A  2  62  B  is  acted  on  by 
its  weight,  represented  by  gg2,  the  cohesion  acting  up  along 
A2j  represented  by  g2w2,  the  reaction  of  the  wall  acting  parallel 
to  Af  and  the  resultant  of  the  normal  reaction  N  of  the  plane 
A 2  and  the  friction  N  tan  <p  acting  up  along  Ai.  The  resultant 
of  N  and  N  tan  #  makes  an  angle  <p  with  the  normal  to  Ai.  It 
thus  has  the  direction  s2  g  or  c2  g. 

The  sides  of  the  closed  polygon  gg2  n^c^g,  thus  represent 
the  four  forces  in  equilibrium  acting  on  the  prism  Ail^B  :  ggz 
to  scale  equaling  its  weight,  g2  n2,  the  cohesion  acting  up  along 
A2,  nzCz,  the  wall  reaction  and  c2g  the  resultant  of  N  and 
N  tan  <p. 

Similarly,  the  sides  of  the  polygons, 

ggi  ni  ci  g,  gg3  ns  c3g,  .  .  ., 
represent  the  forces  acting  on  the  prisms 

AihB,  A$bsB,  .  .  ., 
each  treated  in  turn  as  the  true,  prism  of  rupture. 

As  there  can  be  only  one  true  prism  of  rupture,  it  will  now 
be  proved  that  it  is  the  trial  prism  corresponding  to  the  greatest 


02]  GENERAL   GRAPHICAL  METHOD  203 

ic}  n2  c2  in  this  instance.  This  is  true,  because,  since  w2  is  fixed, 
or  any  less  thrust  than  n%  c2,  the  point  GZ  falls  to  the  left  of  its 
irst  position  and  the  new  c2  g,  representing  the  part  of  the  re- 
iction  of  plane  A  2,  due  to  N  and  N  tan  $  only,  will  thus  make 
t  greater  angle  than  p  with  the  normal  to  A2,  which  is  incon- 
;istent  with  the  laws  of  stability  of  earth.  Since  n2  c2  is  thus 
)roved  to  be  the  true  wall  reaction  (equal  and  opposed  to  the 
iarth  thrust),  the  other  trial  thrusts,  n\  Ci,  ns  c3)  n£C±,  must 
low  be  lengthened  to  equal  w2  c2.  The  new  points  Ci,  c3,  c4, 
vi\\  thus  lie  to  the  right  of  the  old  points,  hence  the  new  gci, 
1C*,  gc±,  will  all  make  angles  less  than  <p  with  the  normals  to  the 
planes  Ai,  A$,  ^4,  so  that  stability  is  everywhere  assured  and 
notion  downward  is  only  impending  along  the  plane  A  2  of  the 
mass  A2  &2  B. 

It  will  be  recalled  that  in  an  indefinite  mass  of  earth,  no 
stress  exists  along  such  planes  as  i&i,  262,  etc.,  which  is  in  agree- 
ment with  the  above  construction.  Similar  conditions  pertain 
to  an  incompressible  earth  and  with  a  rigid  wall  upon  an  in- 
compressible foundation;  but  for  actual  earths  and  walls  on 
compressible  foundations,  there  will  be  perhaps  an  appreciable 
moving  over  of  the  top  of  the  wall.  In  this  case  the  earth,  in  a 
tendency  to  descend,  will  exert  tension  along  2&2,  etc.,  if  it  is 
capable  of  exerting  tension. 

Suppose  a  plane,  as  A2,  is  extended  to  meet  the  free  sur- 
face, say  at  2',  then  if  the  earth,  in  its  impending  descent, 
can  exert  sufficient  tension  along  262  to  drag  down  the  wedge 
262  2',  the  full  cohesion  and  tension  will  be  exerted  along  the 
whole  length  of  the  plane  A  2'  and  similarly  for  other  trial  planes 
of  rupture.  On  dealing  with  the  forces  acting  on  the  wedges 
of  rupture  of  the  type  A  2'  B,  cohesion  acting  along  the  entire 
length  A2f,  the  construction  otherwise  proceeds  as  before  and 
it  necessarily  leads  to  a  less  earth  thrust. 

Although  this  minimum  thrust  may  sometimes  be  exerted, 
it  seems  unwise  to  depend  on  it,  for  after  heavy  rains  and  the 
subsequent  drying  out  of  the  earth,  contraction  occurs  and 
vertical  cracks  often  appear,  particularly  in  clayey  earth;  con- 
sequently tension  along  the  vertical  planes  cannot  be  safely 


204    EARTH   ENDOWED   WITH   BOTH    COHESION   AND   FRICTION 

allowed.  The  construction  as  given  in  Fig.  68  is  therefore  rec- 
ommended. By  this  construction,  for  level-topped  earth  of 
indefinite  extent,  the  thrust  on  a  vertical  plane  Ah  is  horizontal 
and  exactly  equal  to  that  given  by  the  formula  for  E  of  Art.  96. 
When  Ah  =  2  x'  =  2  .b2  2,  the  thrust  is  appreciable,  but  if 
the  second  method  above,  corresponding  to  wedges  of  rupture  of 
the  type  Ai'  h  is  assumed,  it  will  be  found  that  the  thrust  on 
Ah  is  zero.  In  other  words,  when  sufficient  tension  is  exerted 
along  vertical  planes,  as  ibi,  2&2,  .  .  .,  a  vertical  bank  will  be 
self-supporting  when  of  height  ixr\  whereas  if  all  possible  ten- 
sion is  ignored  or  in  case  of  vertical  cracks,  the  bank  is  only 
self-supporting  for  the  height  x'  or  one-half  the  former  height. 
This  is  in  exact  agreement  with  the  results  of  Art.  97. 

EXAMPLE.  In  Fig.  68,  let  Ah  —  10  ft.  and  suppose  the  inner  face  of  the 
wall  AB  to  be  battered  2  ft.  in  10  ft.;  take  i  =  15°,  <p  =  30°,  c  =  100  lb./ft.2, 
w  =  100  Ib.  ft.3  and  find  x'  =  3.46  ft.  and  E  =  thrust  on  AB,  making  the 
angle  <p  with  the  normal  to  AB,  1200  Ibs. 

When  AB,  Fig.  68,  is  vertical  and  the  thrust  is  taken  as  acting 
parallel  to  the  surface,  the  results  by  the  construction  can  be 
compared  with  those  given  by  the  circular  diagram.  Since 
the  construction  of  Fig.  68  assumes  a  plane  surface  of  rupture 
whereas  the  true  surface  of  rupture  is  curved  except  when 
i  =  o,  it  is  to  be  expected  that  the  thrusts  as  found  by  the  two 
methods  will  agree  for  i  =  o,  and  will  differ  more  and  more  as 
the  surface  of  rupture  departs  more  and  more  from  a  plane. 
As  .observed  above,  the  results  for  i  =  o  are  identical.  As  i 
increases,  the  thrusts  by  the  two  methods  remain  practically 
the  same  until  i  is  very  nearly  equal  to  tp.  When  i  <  <p}  the 
true  surface  of  rupture,  AS,  Fig.  60,  is  nearly  plane. 

Thus  for  i  =  15°,  <p  =  30°,  c  =  w  =  100,  the  construction  of  Fig.  68, 
gives  the  thrust  on  Ah  =  43,200  Ibs.,  when  Ah  =  51.75  ft.  The  circular 
diagram  gives  the  unit  thrust  at  this  depth  =  1790  Ib.  ft.2,  it  being  zero  at 
x  =  346ft.  Hence  the  total  thrust  is  >£  1790  X  (51. 75  —  3.46)  =  43,200  Ibs. 

When  i  =  <p,  and  especially  when  i  >  <p,  the  thrusts  by  the 
two  methods  disagree.  In  Fig.  61  where  i  >  tp,  it  is  seen  that 


102,  103]  CENTER    OF   PRESSURE  205 

AS  departs  considerably  from  a  straight  line  and  especially  for 
the  depth  x0  =  BC,  the  thrusts  by  the  two  methods  differ  quite 
appreciably. 

Thus  for  i  =  25°,  <f>  -  20°,  c  =  200  lb./ft.2,  w  =  100  lb./ft.3,  we  have 
x'  =  5.71  ft.,  x0  =  23.8  ft.  and  the  total  active  thrust  on  the  vertical  plane 
x0  in  depth,  by  the  circular  diagram  method  is  14,150  Ibs.,  Art.  94;  whereas, 
by  the  construction  of  Fig.  68,  it  is  found  to  be  12,600  Ibs.,  the  plane  of  rupture 
making  an  angle  of  about  37°  with  the  horizontal. 

With  the  same  data,  the  total  thrust  on  a  vertical  plane  x  =  45.71  ft.  in 
depth,  by  the  construction  of  Fig.  68  is  84,000  Ibs.,  the  plane  of  rupture  mak- 
ing an  angle  of  about  30°  with  the  horizontal.  The  corresponding  unit 
thrust  r'  at  x  =  45.71  (assuming  a  linear  variation  from  x .  =  5.71,  where  it 
is  zero)  is  found  from  the  formula,  tf  r'  X  40  =  84,000  .'.  r'  =  4200  lb./ft.2; 
whereas  by  the  roughly  approximate  method  of  Art.  101,  r'  =  4400  lbs./ft2. 

103.  Center  of  Pressure.  The  center  of  pressure  on  a  wall 
with  the  inner  face  AB,  Fig.  68,  battered,  is  indeterminate.  If 
it  is  assumed  to  vary  uniformly  from  zero  at  F  to  a  maximum 
at  A,  it  will  lie  on  AF  at  one-third  its  length  from  A.  This 
hypo  thesis  may  suffice  for  small  batters;  but  where  AB  has  a 
large  batter,  it  is  safer  to  proceed  as  follows:  resolve  the  thrust 
on  AB  into  vertical  and  horizontal  components  and  assume 
the  horizontal  component  to  act  on  AF  at  1/3  AF  from  A 
and  the  vertical  component  to  act  on  AB  at  1/3  AB  from  A. 
The  vertical  component  is  a  function  of  the  surface  slope,  the 
weight  of  ABh  and  the  friction.  The  friction  at  any  point  is 
due  to  the  normal  pressure  multiplied  by  tan  <?',  and  is  increas- 
ing uniformly  over  BF  (due  to  the  normal  component  of  the 
weight  of  earth  vertically  over  BF)  and  at  a  much  more  rapid 
rate  over  AF.  Considering  the  three  influences,  it  seems  safe 
to  regard  the  center  of  pressure  of  the  vertical  component  to 
act  on  A  B  at  i  /3  A  B  from  A . 

Where  the  wall  leans  toward  the  earth,  or  AB  is  to  the 
right  of  Ah,  the  center  of  pressure  of  the  whole  thrust  can  be 
safely  taken  on  AF  at  1/3  AF  from  A. 

EXAMPLE.  Consider  a  wall  of  .uniform  thickness  5.18  ft.,  leaning  toward 
the  earth,  of  vertical  height  38.2  ft.  and  a  batter  of  1/8  or  i^  inches  to  the 
foot,  the  masonry  weighing  150  lb./ft.3.  Suppose  the  wall  backed  by  level- 
topped  earth  weighing  112  lb./ft.3,  with  c  =  500  lb./ft.2  and  <f>  =  1 8°  30'. 


206    EARTH  ENDOWED   WITH   BOTH   COHESION   AND   FRICTION 

Taking  the  thrust  against  the  wall  as  making  the  angle  i8°3o'  with  the 
normal,  find  by  the  construction  of  Fig.  68,  the  thrust  to  be  11,300  pounds. 
Regarding  this  thrust  as  acting  on  A  F  at  ]/$  AF  above  A  and  combining 
with  the  weight  of  wall,  the  center  of  pressure  on  the  base  (drawn  at  right 
angles  to  the  faces)  is  found  to  pass  0.47  ft.  to  left  of  its  center.  If  the  wall 
has  a  footing  course  7.22  ft.  in  width,  flush  with  the  rear  face,  the  center  of 
pressure  on  its  base  passes  0.53  ft.  to  right  of  its  center. 

The  dimensions  are  those  of  a  wall  built  by  Brunei,  as  given  by  Resal,* 
at  the  entrance  to  the  Mickleton  tunnel,  the  earth  being  blue  lias  clay  whose 
angle  of  repose,  when  humid,  was  only  i8°3O/.  The  wall,  however,  had  a 
projection  on  its  back — not  a  counterfort — and  a  footing  course  and  it  stands 
although  its  thickness  is  less  than  1/6  its  height.  It  is  seen  above  that  its 
stability  is  assured  if  the  coefficient  of  cohesion  is  500  lb./ft.2  Although  the 
weights  of  the  projection  of  the  footing  beyond  the  front  face  and  the  noted 
projection  at  the  rear  face  (which  sustained  the  weight  of  the  earth  over  it) 
were  not  directly  introduced  in  the  computation,  they  were  indirectly  consid- 
ered in  taking  the  weight  of  masonry  at  150  lb./ft.3  in  place  of  146  lb./ft.3, 
the  computed  value. 

If  we  take  c  =  300  lb./ft.2,  the  earth  thrust  is  found  to  be  18,400  Ibs. 
and  the  resultant  on  the  base  of  the  footing  of  width  7.22  ft.,  cuts  the  base 
0.9  foot  from  its  center  or  within  the  middle  third,  so  that  the  wall  is  stable 
with  c  =  300  lb./ft.2,  which  is  not  an  unreasonable  value  for  the  clay  con- 
sidered. 

104.  Braced  Trenches.  As  a  trench  with  vertical  sides  is 
dug,  it  is  assumed  that  the  usual  sheeting,  rangers,  and  bracing 
are  put  in  and  that  the  bracing  is  always  well  keyed  up,  so  as 
to  exert  an  active  pressure  on  the  earth.  The  least  pressure 
necessary  for  stability  can  be  found  from  the  construction, 
Fig.  68,  if  AB,  the  face  of  the  trench  is  taken  vertical,  Bb4  and 
Af,  horizontal,  so  that  n\c\,  n^c^  •  .  .,  are  drawn  horizontally 
to  intersection  with  gsi,  gs2,  ....  The  longest  of  the  corre- 
sponding lines  nc,  to  scale,  gives  the  thrust  that  must  be  exerted 
by  the  braces  to  keep  the  prism  of  rupture  from  descending. 
But,  as  hitherto  pointed  out,  this  least  force  that  the  braces 
must  exert  per  linear  foot  of  trench  is  exactly  given  by  the 

formula, 

.  .  ..  /     0       (p\ 

£  =  K(*-*0'to»>(45°-;), 

2  C 


where,  xf  =  —  tan  (45°  -\ — ). 
ID  V  a/ 


*  Poussee  des  Terres,  Deuxieme  Par  tie,  p.  160. 


104]  BRACED   TRENCHES  207 

. 

As  a  numerical  illustration,  assume  a  trench  40  ft.  deep, 
p  =  33°  41',  c  =  100,  w  —  100  .'.  x'  —  3.74  ft.  and  x  —  x'  = 
36.26  ft.;  hence  the  least  force  the  braces  must  exert  per  linear 
foot  of  trench  is, 

50  (36.26)2  X  0.2866  =  18,844  Ibs. 

From  this  value,  knowing  the  width  of  trench  and  the  spacing 
of  the  bracing,  the  size  of  the  members  can  be  computed. 

It  is  very  likely  that  lower  values  of  <f>  and  much  higher  values 
of  c  can  be  counted  on  for  consolidated  earth. 

As  to  the  next  question  of  the  center  of  pressure,  it  is  evident 
from  the  method  used  in  construction  that  the  distribution  of 
stresses  cannot  be  the  same  as  for  a  retaining  wall.  For  the 
braced  trench,  the  earth  at  first  simply  resists  the  pressure 
exerted  by  the  braces  when  first  put  in  and  keyed  up  tight, 
particularly  on  that  upper  portion  where  the  active  earth  thrust 
is  nothing  or  very  small.  The  digging  is  now  continued  for 
several  additional  feet  without  bracing,  so  that  no  pressure 
can  be  exerted  on  the  unprotected  portion.  Then  sheeting, 
rangers,  and  bracing  are  put  in  to  the  bottom  of  the  trench  so 
far  dug  and  again  the  bracing  is  keyed  up.  The  bottom  braces 
;  at  first  take  only  the  stress  due  to  the  keying  up,  but  as  the 
construction  proceeds  the  braces  take  more  and  more  of  the 
active  earth  thrust,  which  increases  with  the  depth.  In  the 
case  of  a  retaining  wall  the  active  earth  thrust  is  exerted  on 
the  whole  area  below  the  depth  x' ,  but  for  the  braced  trench, 
no  pressure  can  be  exerted  on  an  unsheathed  area  and  the  thrust 
that  would  be  exerted  on  this  area  for  a  retaining  wall  is  carried 
entirely  by  the  bracing  above  it.  It  can  thus  very  well  happen 
that  the  upper  or  middle  braces  receive  more  stress  in  the  end 
than  the  lower  braces.  In  fact,  this  was  asserted  to  be  generally 
true,  for  well-drained  material,  by  many  engineers  in  the  valuable 
discussion  on  Mr.  Meem's  paper,*  though  others  asserted  that 
in  wet  or  saturated  ground,  the  lower  braces  were  most  severely 
stressed. 

If  we  divide  the  value  of  E  above  by  the  depth  of  trench 

"The  Bracing^of  Trenches  and  Tunnels,  with  Practical  Formulas  for 
Earth  Pressures,"  Trans.  Am.  Soc.  C.  E.,  vol.  LX.,  p.  i. 


EARTH  ENDOWED  WITH  BOTH  COHESION  AND  FRICTION 

to  get  the  average  intensity,  it  would  seem  to  be  safe  to  use 
double  this  to  figure  the  intensity  of  stress  at  any  point,  and  to 
design  all  the  braces  for  double  the  average  intensity,  since  the 
exact  distribution  of  stresses  is  unknown. 

As  shown  above,  the  true  value  of  E  is  less  than  that  given 
above  when  no  vertical  cracks  appear  in  the  earth,  which  explains 
why  constructors  deem  it  so  essential  to  get  in  the  bracing, 
well  keyed  up,  as  soon  as  possible  to  prevent  cracks,  which  are 
otherwise  apt  to  form,  particularly  after  heavy  rains. 

105.  Surcharged  Walls,     When  the  surcharge  is  of  the  type 
shown  in  Fig.  69,  lines  ¥2  and  24  are  drawn  x'  ft.  below  and 
parallel  to  Bb%,   bjb^   respectively.     The  weights  of  the  trial 

prisms  of  rupture,  Ai  bi  B,  A2  bz  B,  A$  b3  bz 
B,  etc.,  are  found  and  laid  off  to  scale  from  g, 
Fig.  68,  and  the  construction  proceeds  as  be- 
fore to  find  the  thrust  against  the  wall  AB. 
As  in  the  corresponding  case  of  the  sur- 
charged wall,  Art.  43,  only  with  cohesion 
omitted,  the  center  of  pressure  on  AB, 
Fig.  69,  probably  lies  above  the  point  on  AF,  y$  AF  above  A, 
particularly  for  high  and  steep  surcharges.  If  this  center  is 
taken  0.4  AF  above  A,  it  would  appear  to  be  on  the  safe  side. 
As  the  slope  of  Bbz  diminishes  toward  zero,  the  factor  changes 
gradually  from  0.4  to  y£. 

1 06.  Pressures  on  Tunnel  Linings.    In  Fig.  70  is  shown  a 
vertical  transverse  section  of  a  tunnel  of  width  b,  length  I,  with 
the  earth  extending  to  a  height  h  above  it.     If  this  tunnel  has 
been  driven  by  the  use  of  shield  or  poling  boards,  the  earth 
will  tend  to  settle  over  it,  and  part  of  the  weight  of  this  earth 
directly  over  the  tunnel  will  be  sustained  b>;  the  cohesion  and 
friction   (resulting  from  the  lateral  thrust)   exerted  along  the 
sides  and  acting  vertically  upwards. 

At  the  depth  y  below  the  surface,  consider  the  conditions  of 
equilibrium  of  a  horizontal  lamina  of  depth  dy,  width  b,  and 
length  /. 

Let  V  =  the  vertical, unit  pressure  at  depth  y. 
L  —  the  horizontal  unit  pressure  at  depth  y. 


106] 


PRESSURES   ON  TUNNEL  LININGS 


209 


w  =  weight  of  earth  per  cubic  unit. 

(p  =  angle  of  friction,  /z  =  tan  <p. 

c  =  coefficient  of  cohesion. 

A  =  b  I  =  area  of  a  horizontal  section  of  tunnel. 

U  =  perimeter  of  section  as  defined  below. 
Thus,  when  the  ends  as  well  as  the  sides  are  considered  as 
offering  support  to  the  earth  over  the  tunnel,   U  =  2  b  +  2  /; 
for  a  very  long  tunnel  of  length  /,  let  U  =  2  I',  but  for  a  short 

section  under  construction,  for  which  I  =  — ,  say,  the  two  sides 

2 

and  one  (the  front)  end  alone  will  be  considered  as  offering  a 
support,  so  that  U=2l  +  b  =  2b  =  4l.    The  first  supposi- 


i_3 


dy 


(V+AW 


Tunnel 


\jjdy 


FIG.  70 

tion,  U  =  2  (b  -\- 1),  is  identical  with  that  pertaining  to  the 
ordinary  rectangular  bin.  In  fact  the  theory  that  follows  is 
only  a  slight  modification  of  Janssen's  bin  theory. 

For  any  one  of  the  three  cases,  the  total  lateral  force  acting 
on  the  lamina  is  Z,  U  d  y  and  the  f rictional  and  cohesive  forces 
acting  vertically  upwards  on  the  sides  and  ends  are  L  U  dy  /* 
and  U  dy  c,  as  marked  on  the  figure.  Hence,  if  V  +  AF  is 
the  unit  pressure  on  the  under  side  of  the  lamina,  acting  ver- 
tically upwards,  we  have  for  equilibrium 

(V  +  AF)  A  --  VA  -  wAdy  +  (L  U»  +  Uc)  dy  =  o 

One  approximation  has  been  introduced  here,  since  for  a 

depth  yf  =  -—  tan  [45°  +  — ),  L  is  zero  for  cohesive  earth,  so 

W  \  2  / 


210    EARTH   ENDOWED   WITH   BOTH   COHESION   AND   FRICTION 

that  the  term  (L  Un  +  Uc)  dy  reduces  to  Uc  dy,  for  this  entire 
depth  from  the  surface;  but  for  large  values  of  y  the  difference 
in  final  results  will  be  inappreciable.  Another  approximation 
consists  in  placing, 

L=  Vk 

where  k  is  a  constant  to  be  determined  by  experiment.  Many 
experiments  on  pressures  in  bins  filled  with  grain,  show  that  k 
is  not  a  constant,  still  the  supposition  leads  to  fairly  good  practical 
results. 

On  substituting  L  =  Vk  in  the  equation  above  and  solving 
for  d  F  (put  for  AF), 

(wA-cU-  k»UV)dy 

~ 


wA-cU-kpUV  A 

On  integrating, 

^M— 

Oge  C'  — A 

When  y  =  o,   V  =  o  .'.  C  =  loge  (w  A  —  c  U) 


w  A  —  c  U 

Putting  y  =  h,  then  V  (hereafter)  will  denote  the  vertical 
unit  pressure  at  depth  h,  or  at  the  roof  of  the  tunnel, 
w  A  —  c  U  —  kiiU  V 


=  c     A 

w  A  —  c  U 

where  e  =  2.71828  .  .  .  ,is  the  Napierian  base. 

L=  Vk (2) 

The  formulas  (i)  and  (2),  give  the  values  of  the  vertical 
and  lateral  unit  pressures  V  and  L,  at  the  top  of  the  tunnel, 
at  the  depth  h  below  the  free  surface.  The  earth  over  the 
tunnel  (or  grain  in  a  bin)  will  probably  arch  itself  and  it  can  be 
conceived  to  form  a  series  of  arches  (or  domes)  superposed  on 


106]  PRESSURES   ON  TUNNEL  LININGS  211 

each  other.  If  each  arch  has  the  vertical  thickness  dy,  it  will 
have  the  same  volume  and  weight  Awdy  as  the  horizontal 
lamina  of  Fig.  70,  by  a  theorem  of  the  calculus.  Further,  if  the 
reactions  of  the  sides  on  the  lamina  marked  on  Fig.  70,  be  as- 
sumed to  be  those  of  the  corresponding  arch,  their  resultant  may 
be  supposed  to  represent  the  thrust  of  the  arch  at  the  sides. 
Similarly  for  the  dome.  It  is  thus  seen  that  the  horizontal 
lamina,  having  the  same  reactions  as  the  corresponding  arch 
(or  dome),  can  be  substituted  for  it  in  stating  the  equilibrium 
equations  above.  The  value  of  V  given  by  (i)  is  the  average 
value  at  depth  h  or  at  the  top  of  the  tunnel. 

Experiments  on  grain  bins  lead  to  the  inference  that  the 
pressure  at  the  middle  of  the  roof  is  greater  than  that  near  the 
sides,  but  no  law  of  variation  can  be  stated.  Similarly  the 
lateral  unit  pressure  L  =  kV,  at  the  level  of  the  roof  can  be 
found,  but  the  variation  of  the  lateral  pressure  over  the  vertical 
sides  of  the  tunnel  lining  cannot  be  formulated,  since  so  much 
of  the  weight  of  the  earth  directly  over  the  tunnel,  has  been 
transferred  by  a  kind  of  arch  action  to  the  sides.  It  is  scarcely 
necessary  to  remark  that  the  formula  does  not  give  the  lateral 
pressure  on  the  tunnel  lining,  caused  by  the  swelling  of  clays 
saturated  with  water.  The  results  refer  to  ordinary  ground, 
either  dry  or  moderately  humid  or  to  such  earths  as  do  not 
swell.  The  case  of  saturated  earth  will  be  discussed  later. 

As  a  numerical  application,  consider  a  long  tunnel,  where 
the  ends  are  supposed  to  offer  no  support  to  the  earth  directly 
over  the  tunnel,  so  that  U  =  2  /.  If  the  width  of  tunnel  is  15 

^       A       bl        b 
ft,  then  —  =  —  =  -=  7.5  ft. 

U  2l  2 


Let  <p  =  30°  .'.  M  =  tan  <p  =  0.577,  anc^  assume  w  =  90 
lb./ft.3,  c  =  100  lb./ft.2.  As  to  the  value  to  assume  for  k,  no 
satisfactory  experiments  can  as  yet  be  appealed  to.  For  an  un- 
limited mass  of  earth  devoid  of  cohesion,  with  level  surface, 

L/V  =  tan2  (45°  --  —  )  ,  but  this  value  does  not  agree  generally 

2  ' 

with  the   experimental   determinations   on   large   bins.     Thus, 


212    EARTH   ENDOWED   WITH   BOTH   COHESION   AND   FRICTION 


Jamieson  found  from  experiments  on  a  wooden  bin,  about 
12  by  12  ft.,  filled  with  wheat  (p  =  28°),  k  =  0.60,  whereas, 
tan1  (45°—  <f>/2)'=  0.361.  The  experimental  value  is  10/6  of 
this,  so  that  from  lack  of  definite  data,  there  will  be  assumed 
for  this  value  of  <p, 


L        10 

k  =  —  =  —tan2 
V         6 


For  <p  =  30°,   this  gives  k  =  0.555   and 
mula  (i)  thus  reduces  to, 


=  0.320.     For- 


V  =  1790  (i  - 


A 


0.0427  h  I  ' 


V  is  expressed  in  pounds  per  square  foot,  h  in  feet.* 

The  following  table  gives  the  values  of  V,  corresponding  to 
assumed  Values  of  h.  The  results  are  shown  graphically  in 
Fig.  71.  By  the  assumption,  for  any  V,  L  =  kV  =  0.555  V. 


h 

V 

h 

V 

h 

V 

Ft. 

Lb./Ft.2 

Ft. 

Lb./Ft.2 

Ft. 

Lb./Ft.2 

10 

621 

40 

1,466 

70 

I,7OO 

20 

1,022 

50 

1,580 

80 

1,730 

30 

1,292 

60 

1,650, 

100 

1,765 

For  large  values  of  h,  e~klJ"A:h 


is  very  small,  and  as  h  in- 
creases indefinitely,  eq.  (i)  approaches  the  limit, 

i  r~     A        ~i 
Um  V  =  —  \  w  —  -  c  \     .      .      .      .      (3) 

K/JL1-          U  -J 

which  in  the  present  example,  amounts  to  1790  lb./ft.2,  a  value 
which  differs  but  little  from  that  at  h  =  ioo'fc.  This  limiting 
value  of  V  can  then  be  regarded  as  practically  the  maximum 
value  of  V  for  large  values  of  h.  The  expression  for  this  limiting 
value  of  V  —  call  it  Vm  —  can  easily  be  obtained  independently. 


*  A  short  table  of  Napierian  logarithms  is  a  convenience  in  computation. 


Thus  when  h 
=  loge  3.60. 


30  ft.,  e°-°427h  =  e1'2**  =  3.60,  since  from  the  Table,  1.281 


106] 


PRESSURES    ON   TUNNEL   LININGS 


213 


'20 


£40 


100 


Thus  in  Fig.  70,  considering  the  whole  series  of  horizontal  lamina, 
it  is  plain  that  the  greatest  value  of  V  would  be  realized  if  the 
weight  of  any  lamina  was  entirely  sustained  by  the  friction  on 
the  sides  (induced  by  the  lateral  thrust)  and  the  cohesion  acting 
there,  for  then  this  same  condition  of  affairs  would  exist  for 
all  the  lower  lamina,  Vm  being  transmitted  vertically  down- 
ward, unchanged  to  the  tunnel. 
Stating  this  condition  in  algebraic 
form, 

Aw  dy  =  (L  UJJL  +  Uc)  dy. 

Putting  L  =  Vmk,  this  equa- 
tion reduces  to  (3)  above.  The 
equation  is  almost  exactly  true  for 
large  values  of  y  and  is  practically 
true  for  much  smaller  values,  as 
may  be  seen  from  the  results  above 
or  from  Fig.  71. 

Recurring  to  eq.  (i),  it  is  seen  that  when  c  is  large  enough, 

w c     may  be  zero  or  negative.     In  either  case,  it  will 

be  shown  that  the  earth  over  the  tunnel  can  be  held  up  by  the 
cohesion  acting  along  the  vertical  sides,  so  that  V  must  be 
placed  equal  to  zero.  Thus  when  U  is  the  perimeter  pertaining 
to  the  particular  case,  the  total  cohesive  force  that  can  be 
exerted  for  the  height  h  is,  c  Uh.  The  weight  of  the  vertical 
prism  of  earth  directly  over  the  tunnel  is  wAh,  so  that  the 
cohesion  alone  can  support  the  weight  of  the  prism  when  cUh>_ 

wAh  or  when  c  _>.  w  — ,  as  stated. 
U 


4  8  12         16 

Hundred  Pounds 


FIG.  71 


From  the  three  equations, 
A_ 
U  ' 


k  =  T 
6 


the  values  of  Vm  and  L,  have  been  computed  for  w  =  90  lb./ft.3 
and  various  values  of  c,  b,  $  and  A/U,  and  inserted  in  the 
following  table: 


214    EARTH  ENDOWED   WITH   BOTH   COHESION   AND   FRICTION 


c 

b 

Vm  in  Lb./Ft.2 

L  in  Lb./Ft.2 

Lb./Ft.2, 

Ft. 

<f> 

A    b 

A   b 

A   b 

A   b 

U   4 

U    2 

U  ~  4 

U  ~7 

(i) 

(2) 

(3) 

(4) 

(5) 

(6) 

(7) 

100 

15 

30° 

740 

1,790 

410 

1,000 

100 

30 

30 

1,790 

3,900 

1,000 

2,160 

IOO 

15 

45 

830 

2,010 

240 

570 

IOO 

30 

45 

2,OIO 

4,370 

570 

1,250 

400 

15 

30 

0 

860 

0 

480 

400 

30 

30 

860 

2,960 

480 

1,640 

400 

15 

45 

O 

960 

0 

270 

400 

30 

45 

960 

3,320 

270 

950 

The  pressures  given  in  columns  5  and  7  refer  to  a  long  tunnel 
as  previously  explained.  For  a  short  length  of  tunnel  about 

b 
the  heading,  /  was  arbitrarily  taken  =  — ,  and  the  earth  directly 

2 

over  this  length  of  tunnel  was  supposed  partially  supported 
by  the  friction  and  cohesion  acting  on  the  two  sides  and  the 
front  end,  extended  to  the  surface,  whence  U '  =  2  /  +  b  =  26, 
A  =  b2/2,  therefore  A/U  =  6/4-  The  corresponding  pressures 
are  given  in  columns  (4)  and  (6) . 

The  values  for  c  =  100  are  intended  to  apply  to  soft  ground; 
those  for  c  =  400,  to  hard  consolidated  earth.  In  both  cases, 
the  earth  is  supposed  to  be  dry  or  moist  but  not  saturated 
with  water.  For  c  =  400  lb./ft.2,  it  is  observed  that  the  average 
pressures  are  both  zero  at  the  working  faces  of  the  15  ft.  tunnel. 
For  very  hard  ground,  there  is  but  little  doubt  that  c  exceeds 
considerably  400  lb./ft.2  or  that  the  pressures  will  be  smaller 
than  any  given.  In  fact  only  extended  experimental  deter- 
minations of  the  coefficient  c  and  the  angle  ,of  friction  <p  after 
the  method  suggested  in  Chap.  I,  can  lead  to  satisfactory  results 
in  all  cases  where  c  and  <p  enter  as  determining  factors.  Mr. 
J.  C.  Meem,  M.  Am.  Soc.  C.  E.,  in  a  paper  on  "Pressure,  Re- 
sistance and  Stability  of  Earth,"*  gives  a  table  of  pressures  on 
tunnel  linings,  which  may  be  compared  with  the  above. 


*  Trans.  Am.  Soc.  C.  £.,  vol.  LXX,  p.  387. 


107] 


TUNNELS   IN    SATURATED   EARTH 


215 


Water  Surface 


107.  Saturated  Earth.  When  the  earth  is  completely  satu- 
rated with  water,  the  buoyant  effort  of  the  water  must  be 
considered,  but  there  is  by  no  means  agreement  as  to  how  this 
shall  be  done.  If  the  earth  particles  are  to  be  treated  as  so 
many  marbles,  so  that  the  water  has  free  access  to  any  horizontal 
or  vertical  plane,  the  procedure  is  simple.  Thus,  if  a  cubic  foot 
of  earth  with  40%  voids,  weighs  in  air  90  Ibs.,  it  contains  0.6 
cu.  ft.  of  solids  and  the  buoyant  effort  of  the  water  equals  the 
weight  of  an  equal  volume  of  water,  or  0.6  X  62.5  =  37.5  Ibs.;  so 
that  the  weight  of  the  cubic  foot  of  earth  in  water  is  90  —37.5 
=  5  2 . 5  Ibs.  If  for  any  earth,  the  weight  of  a  cubic  foot  of  the  earth 
in  water  as  determined  in  this  way,  is  called  w1 ',  then  wf  is  to 
replace  w  in  the  formulas  above  and  the  constants  c  and  <p  to  be 
used  must  be  determined  by  experiment  for  the  saturated  earth. 

Let  h0,  Fig.  72,  represent  the  depth  from  the  water  surface 
to  the  top  of  the  tunnel.  Then,  when  h0  =  h,  on  replacing 
w.by  w'  and  using  the  values  of  c  and 
<p  for  saturated  earth,  formulas  (i)  and 
(2)  will  give  V  and  L  for  the  earth  at  the 
top  of  the  tunnel.  To  these  values  add 
62.5  h0  to  get  the  true  pressures  due  to 
both  earth  and  water.  When  h0  <  h, 
rough  average  values  of  w  and  w',  c  and 
V  may  be  used  in  getting  an  approxi- 
mate value  of  the  earth  pressure. 
The  values  of  w',  c  and  <p  used  will  evidently  depend  upon 
the  ratio  h0/h. 

When  k0  >  h,  lower  and  upper  limits  can  be  found  as  in  the 
following  numerical  example,  referring  to  a  tunnel  under  a  river 
through  clayey  silt,  for  which  say  <p  =  10°,  c  =  200  lb./ft.2, 
w  =  90  lb./ft.3  .".  as  above,  w'  =  52.5.  Assume,  k  =  tan2 

(45° -J  =  0.704  .'.  kfj.  =  ktan  10°  =  0.124.    Let/*  =  40 ft., 

h0  =  60  ft.  and  for  a  long  tunnel  b  =  15  ft.  wide,  A/U  =  7.5. 
Eq.  (i),  Art.  106,  gives  the  vertical  pressure,  for  h  =  40  ft., 
due  to  the  earth  alone, 


=~  —  =  —  = 

,  — 

1 

jfc 

i 

Saturated 
Earth 
«—_  &  +. 

h, 

l_^ 

///////^^//^/^//// 

£    Tunnel    $ 
L-zzzamB   ? 

FIG.  72 


216    EARTH   ENDOWED   WITH   BOTH   COHESION   AND   FRICTION 

V  =  1564  (i  -  e-'ol66h)  =  1564  X  0.485  =  760  lb./ft.2 
This  is  less  than  the  true  pressure,  as  the  weight  of  water  above 
the  earth  must  cause  some  additional  pressure  in  the  earth. 

On  adding  the  full  water  pressure  for  the  height  ^,62.5  X  60 
=  375°)  we  find  the  total,  4510  lb./ft.2  Similarly  we  add  3750 
to  k  V  to  find  the  lateral  unit  pressure.  An  upper  limit  for  the 
pressure  corresponds  to  h  indefinitely  large,  whence  V  —  Vm  = 
1564.  To  this  add  the  water  pressure,  3750,  giving  the  total 
5314  lb./ft.2  The  true  vertical  unit  pressure  thus  lies  between 
4510  and  5314  lb./ft.2  and  the  true  lateral  pressure  between 
(3750  +  k  V)  and  (3750  +  k  VJ  lb./ft.2 

For  small  values  of  <p,  it  seems  more  reasonable  to  assume 

k  =  tan2  (45°— — J,  in  place  of  10/6,  this  amount  as  before, 

since  k  is  probably  always  less  than  unity      In  fact,  for  <p  =  o 
(water  pressure)  k  is  i,  so  that  whatever  factor  is  applied  to 

tan2  (  45°  —  —  J  for  the  larger  values  of  <p,  it  must  gradually 

approach  i  as  <p  approaches  zero. 

The  above  solution  may  suffice  for  gravel,  but  for  sand  or 
silt  with  perhaps  much  clayey  matter,  the  pores  are  more  or  less 
clogged  up  and  there  is  perhaps  intimate  contact  of  a  part  of 
the  earth  with  the  roof  of  the  tunnel,  so  that  the  water  cannot 
get  under  it  to  produce  a  lifting  effect,  and  the  buoyant  effort 
of  the  water  is  much  diminished.  Perhaps  this  may  be  more 
marked  under  small  heads  than  large  ones.  Before  any  definite 
percentage  of  reduction  can  be  stated,  experiments  on  a  large 
scale  and  on  every  kind  of  material  usually  met  with,  should 
be  made;  but  as  a  mere  numerical  illustration,  Assume  as  found  in 
Mr.  Meem's  experiment*  that  the  water  pressure  through  sand 
having  40%  voids  is  diminished  40%. 

With  the  data  of  the  preceding  example,  the  weight  of  the 
cubic  foot  of  earth  in  water  is  now,  w"  —  go  —  (0.4  X  0.6 
X  62.5)  =  75  pounds.  On  replacing  w  by  w"  in  formula  (i), 

*  Art.  9,  foot-note. 


107]  TUNNELS   IN   SATURATED   EARTH  217 

we  find  V  —  1416  lb./ft.2  To  this  add  the  water  pressure,  0.4 
(62.5  X  60)  =  1500,  and  the  lower  limit  to  the  vertical  pressure 
at  the  top  of  the  tunnel  is  found  to  be  2916  lb./ft.2  The  upper 
limit  is, 

Vm  +  1500  =  2918  +  1500  =  4418  Ib.  /ft.2 
Similarly  the  lower  and  upper  limits  of  L  at  the  roof  of  the 
tunnel  are  (1500  +  k  V)  and  (1500  +  k  Vm)  respectively.* 

*  At  first  sight,  it  might  be  thought  that  a  definite  solution  of  the  case 
where  h0  >  h,  the  weight  of  water  over  the  earth  being  62.5  (h0  —  h)  lb./ft.2  = 
V  say,  could  be  made  by  regarding  the  water  as  earth  of  equal  weight  and 
determining  the  constant  C,  Art.  106,  by  making  V  =  V  when  y  =  o. 

-kv-V-h 
The  resulting  value  of  Fat  the  tunnel  is  greater  than  before  by  V  e          A    ; 

but  if  to  this  is  added  the  remaining  water  pressure  62.5  h,  the  result  will  be 
found  to  be  less  than  the  lower  limit  in  the  first  solution  above.  The  reason 
is  that  the  theory  implies  transference  of  a  certain  portion  of  the  weight  of 
water  over  the  earth  to  the  sides,  which  is  inadmissible. 


CHAPTER  VI 

THEORY  PERTAINING  TO  BOTH  DEEP  AND  SHALLOW  BINS 

1 08.  A  shallow  bin  may  be  defined  as  a  bin  where  the  surface 
of  rupture  cuts  the  free  surface  of  the  filling  or  contained  material. 
Thus  the  ordinary  hopper  bin,  containing  coal,  coke  or  ore,  is 
usually  a  shallow  bin. 

A  deep  bin  is  one  having  a  greater  depth  than  the  highest 
shallow  bin  for  the  same  diameter,  as  in  the  case  of  grain  bins. 
The  theory  pertaining  to  deep  and  shallow  bins  is  entirely 
different,  as  may  be  surmised  from  what  has  preceded. 

Grain  Bins.  The  theory  of  Art.  106  applies  to  deep  vertical 
bins  filled  with  grain,  on  replacing  ju  by  //,  the  coefficient  of 
friction  of  the  grain  on  the  wall  and  (for  simplicity)  making 
c  =  o.  The  formulas  for  V  and  L,  now  take  the  simpler  forms, 


w     A 
For  large  values  of  h,  we  have  practically,    V  =  — - .  — , 

kfJL          U 

w     A 
L  =  —  .  — ;  so  that  L  is  independent  of  k.     This  is  important, 

//       U 

since  for  wheat  alone,  k  has  been  found  by  different  experimenters 
to  vary  from  0.3  to  0.67.     For  a  rectangular  bin,  since  all  four 

A  bl 

vertical  sides  offer  support,  —  =  ,  which  reduces  to  6/4 

U       2  (b  +  /) 

for  a  square  bin.     For  a  circular  section  of  diameter  D,  A  = 
7rZ)2/4,  U  =  irD  :.  A IV  =  D/4. 

The  stresses  V  and  L  are  thus  the  same  for  a  square  bin  and  a 
circular  one  when  b  =  D,  or  when  the  diameters  are  the  same. 

218 


108]  GRAIN  BINS  219 

Janssen,  who  first  deduced  the  above  formulas  for  bins,  found 
by  experiments  on  bins  of  smooth  cribbed  boards, 

k  =  0.67,  //  =  0.3  .".  kfjf  =  0.20. 
The  bins  contained  wheat  weighing  50  lb./ft.3 

Jamieson*  in  an  elaborate  series  of  experiments  on  full  sized 
bins,  found  for  wheat,  w  =  50,  <f>  =  28°,  k  =  0.60,  //  =  0.44, 
kf/  —  0.264,  the  bin  being  of  timber  crib  construction.  A  bin 
12"  square  and  6'  6"  deep,  of  steel  trough  plate,  was  also  filled 
successively  with  grain  weighing  in  lb./ft.3:  for  wheat  50,  peas 
50,  corn  56  and  flax-seed  41.5.  It  was  found  that  the  pressures 
in  the  case  of  the  corn  were  about  the  same  as  for  the  wheat; 
for  the  peas,  about  20%  greater  and  for  the  flax-seed,  10  to  12% 
greater  than  in  the  case  of  the  wheat. 

It  is  possible  that  the  values  of  k  given  above  are  too  high, 
since  experiments  by  Bovey,  Lufft,  Ketchum  and  Pleissner, 
give  values  of  k  varying  from  0.3  to  0.6.  Thus  Pleissner  gives 
for  wheat  in  a  small  plank  bin,  //  =  0.25,  k  =  0.34,  kp  =  0.085, 
and  in  a  rough  concrete  bin,  //  =  0.71,  k  =  0.30,  £//  =  0.213. 
For  a  cribbed  bin,  k  was  found  to  vary  from  0.4  to  0.5.  The 
values  of  k  for  rye  for  the  various  cases,  were  roughly  about  % 
less  and  the  values  of  kf  about  %  greater.  Jamieson  found  for 
wheat  on  steel  flat  plate  riveted,  //  —  0.375  to  0.4;  on  steel 
cylinder  riveted,  //  =  0.365  to  0.375;  on  concrete,  smooth  to 
rough,  ju'  =  0.4  to  0.425  and  on  a  cribbed  wooden  bin,  //  =  0.42 
to  0.45. 

Some  of  the  quantities  given  above  have  been  taken  from 
Ketchum's  "  WaUs,  Bins  and  Grain  Elevators,"  to  which  the 
reader  is  referred  for  an  exceptionally  full  treatment  of  the  sub- 
ject. It  is  found  from  the  experiments  that  the  lateral  pressure 
of  the  grain  on  the  bin  walls  increases  very  little  after  a  depth 
of  3  times  the  width  or  diameter  of  the  bin  is  at  tamed,  and  that 
the  pressures  computed  by  the  formulas  above  agree  closely 
with  the  observed  pressures.  The  curves  of  pressure  shown  in 
Fig.  71  are  similar  to  those  referring  to  grain  bins.  The  pres- 
sures corresponding  to  grain  running  out  of  the  discharge  gates 

*  Trans.  Can.  Soc.  C.  E.,  vol.  XVII,  1903;  Eng.  News,  vol.  51,  1904,  p.  236. 


220        THEORY    PERTAINING   TO   DEEP   AND    SHALLOW   BINS 

rarely  exceeds  by  10%  those  due  to  the  grain  at  rest.  The 
gates  should  be  located  near  the  center  of  the  bin,  for  if  located 
in  the  sides,  the  lateral  pressure  on  the  side  opposite  the  gate 
is  very  materially  increased. 

As  an  illustration  of  the  application  of  the  formulas,  consider  a  square 
wooden  cribbed  bin,  b  =  10'  wide  and  h  =  70'  high,  filled  with  wheat  weighing 
w  =  50  lb./ft.3  Assume  k  =  0.5,  //  =  0.42  .*.  kp  =  0.21.  For  the  square 
bin,  A/U  =  ft/4  =  10/4  and  for  h/b  =  7,  it  is  sufficiently  near  to  compute 
V  from  the  formula, 

w    A         50  X  10 

V™    =   IT~ '  Tr    =  -    =  600  lb./ft.2 

k  n    U       0.21  X  4 

The  total  pressure  on  the  bottom  is  thus,  600  A  pounds,  and  since  the 
weight  of  grain  in  the  bin  is  70  X  5  o  X  A  =  3500  A  pounds,  the  weight  of 
grain  carried  by  the  sides  through  friction  is  2900  A  pounds.  Thus  the  sides 
of  the  bin  carry  29/35  or  71.5%  of  the  whole  weight  of  grain.  This  amount  is 
nearly  unvarying,  even  when  the  grain  is  running  out,  and  incidentally,  it  is 
proof  that  the  friction  of  the  earth  on  a  retaining  wall  is  permanent,  the 
conditions  as  to  moisture  being  the  same  and  that  this  friction  is  exerted  at 
all  times  and  not  simply  at  the  moment  of  failure,  as  is  often  contended. 

The  lateral  pressure  per  square  foot  of  bin  wall,  at  the  bottom,  is,  k  Vm  = 
0.5  X  600  =  300  pounds. 

As  another  illustration,  consider  a  smooth  circular  steel  bin,  25  ft.  diame- 
ter and  75  ft.  high,  filled  with  wheat.  Assume,  w  =  50,  k  =  0.5,  /  =  0.37 
.'.  V  =  0.185.  We.  have  A/U  =  25/4;  hence  substituting  in  the  formula 
for  V  above, 


V  =  1700  \  i  -  — ). 

Since  e2'2  =  9.25,  V  =  1520  lb./ft.2 

The  total  pressure  in  pounds  on  a  horizontal  section  75  ft.  below  the 
surface  of  the  grain  is  thus  1520  A  ;  the  weight  of  grain  in  the  bin  is  75  X  50  A 
—  375°  A ;  so  that  2230  A  Ibs.  is  carried  by  friction  by  the  sides,  where 
A  =  TT  £>2/4.  On  dividing  this  by  the  perimeter  n  D,  the  amount  carried  by 
a  vertical  strip  I  foot  wide  and  75  feet  high  is  found  to  be,  2230  X  25/4  = 
I3>925  pounds.  This  amount  added  to  the  weight  of  the  steel  strip,  is  the 
shear  at  depth  75  ft.  to  be  carried  by  the  horizontal  rivets  for  each  linear 
foot  of  perimeter. 

The  lateral  pressure  at  depth  75  ft.,  is  L  =  k  V  =  0.5  X  1520  =  760 
lb./ft.2 

Similarly  the  vertical  and  lateral  pressures  at  any  depth  can  be  found 
from  the  formulas,  by  substituting  for  h  the  desired  depth.  Or  preferably, 
y  can  replace  h  in  the  formulas,  to  give  V  and  L  at  the  variable  depth  y. 

As  noticed  above,  the  value  of  k  is  the  most  uncertain  quantity  to  be 
assumed  and  the  numerical  results  just  found  will  be  somewhat  altered  if 
k  is  changed  in  value.  Thus,  if  after  Jamieson,  \ve  take  k  =  0.6,  the  other 
constants  being  unaltered,  it  is  found  at  the  depth  h  =  75  ft., 


108,  109] 


SHALLOW   BINS 


221 


V  =  1256  lb./ft.2;  L  =  0.6  V  =  754  lb./ft.2 

The  weight  of  grain  carried  through  friction  to  the  vertical  sides  of  the 
bin  is  now,  (3570  —  1256)  A  Ibs.  and  the  weight  per  lineal  foot  of  circum- 
ference is  15,590  Ibs.  This  weight,  transferred  through  friction,  acts  on  the 
the  inner  side  of  the  bin  wall,  and  this  eccentricity  should  be  considered  in 
designing  any  portion  of  the  wall  as  a  column,  particularly  in  the  case  of 
thick  \vooden  walls. 

109.  Shallow  Bins.  The  case  of  the  shallow  hopper  bin, 
filled  with  coal  or  other  granular  material,  will  alone  be  con- 
sidered in  what  follows.  The  strict  theory  pertaining  to  the 
shallow  bin  is  doubtless  a  very  complicated  one  and  it  may  be 
mentioned  at  the  outset,  that  any  approximate  solutions  in- 
dicated below,  based  upon  a  plane  (and  not  a  curved)  surface 
of  rupture,  are  open  to  objections,  which  will  be  clearly  indicated. 

A  simple  solution  for  ordinary  symmetrical  bins  will,  how- 
ever, be  suggested  that  is  believed  to  be  on  the  side  of  safety. 

To  estimate  the  thrust  on  the  sides  of  the  bin,  it  is  necessary 
to  know  the  weight  per  cu.  ft.  (w)  of  the  filling,  its  angle  of 
friction  (<p)  and  the  angle  of  friction  (<pf)  of  the  filling  on  the 
bin  walls. 

The  following  table  gives  values  of  w,  <p  and  $  for  various 
materials : 


</>' 

Material 

Lb./Ft.3 

4> 

Steel 
Plate 

Wood 
Cribbed 

Concrete 

Bituminous  coal.  . 
Anthracite  coal.  .  . 
Sand 

46-56 
47-58 

QO— 

35° 
27 

T.A  — 

18° 
16 
18 

35° 
25 

70 

35° 

27 

7Q 

Ashes  . 

4O— 

4.0—4.5 

71 

4O 

4O 

Ore  

125— 

T.5-45 

Coke,  piled  loose  . 

23-32 

25 

40 

40 

There  are  three  fundamental  conditions  to  which  the  theory 
must  conform: 

(i)  In  the  symmetrical  bin,  Fig.  73,  symmetrically  loaded, 
whether  the  filling  is  level  at  top  or  heaped,  the  thrust  on  the 
vertical  plane  CW,  must  be  horizontal  from  considerations  of 
symmetry. 


222         THEORY   PERTAINING   TO   DEEP   AND   SHALLOW   BINS 

(2)  The   thrust  on  the  vertical  plane  AB  will  make  the 
angle  </>'  with  the  normal  to  the  plane  and  this  direction  does 
^  not  change,  exactly  as  experiment  shows  to 

\          be  the  case  in  deep  bins. 

(3)  The  thrust  on  BC  makes  an  angle 
with  the  normal  to  BC  which  cannot  ex- 
ceed v  or  equilibrium  will  be  impossible. 
no.  Hopper  Bin.     Coal  Heaped.     The 
bin  selected  for  computation  is  shown  in 
cross-section,   ABCFGH,  Fig.    74,  with- 
out the  interior  bracing.     It  is   the  same 
bin  that  figured  in  the    rather  extended 
discussion  given  in  Engineering  News  in  the  years    1897-98.* 


FIG.  73. 


FIG.  74. 

As  in  the  original  discussion,  the  values  w  =  58,    <p  =  30°,   <pf 
=  15°    will    be    used   below,    though   for  anthracite   coal,   w 


*  For  a  general  discussion  of  bins,  based  on  the  author's  formulas  for  earth 
pressure,  see  article  by  R.  W.  Dull  in  Eng.  News,  of  July  21,  1904. 


110]  HOPPER  BIN.      COAL  HEAPED  223 

=  52,  <p  =  27°,  <?'  =  1 6',  are  regarded  as  nearer  the  true 
values. 

In  this  bin,  AB  =  8',  Ag  =  n',  Cg  =  19.5',  CF  =  10'  and 
the  pocket  is  17'  long.  The  influence  of  the  friction  at  the  ends 
in  diminishing  the  thrust  will  be  ignored.  The  analysis  below 
refers,  as  usual,  to  i  foot  length  of  bin  and  the  coal  will  be 
supposed  heaped  at  the  angle  of  repose,  as  shown  in  the  figure. 

It  is  easy  to  find  the  thrust  E  on  the  vertical  plane  AB, 
the  thrust  being  inclined  <p'  =  15°  to  its  normal,  by  the  graphical 
method,  Art.  39,  Fig.  19,  for  the  surcharge  AKH.  Thus  draw 
APllBK,  to  intersection  P  with  HK  produced.  From  P 
draw  P0f  making  the  angle,  <p  +  </  =  45°  with  AB  or  with 
the  vertical,  to  intersection  Of  with  BOT ,  drawn  parallel  to  AK 
or  at  the  natural  slope.  Suppose  B  Or  produced  meets  KH 
at  D'.  Using  the  mid-point  of  O'D'  as  center,  describe  semi- 
circle through  Of  and  D'  and  lay  off  BIr  equal  to  length  of 
tangent  from  B  to  this  semi-circle.  From  /',  draw  line  parallel 
to  P  0'  to  intersection  Q'  with  KH  and  lay  off  /' ' Lr  =  7'(X; 
then  the  thrust  E  =  w  X  area  shaded,  triangle  -L'l'Q'  =  58 
XK  (64  X  6.2)  =  1 150  Ibs.* 

The  plane  of  rupture  is  B  (X;  whereas  if  the  slope  AK  was 
unlimited,  the  plane  of  rupture  would  be  B  D'  extended  in- 
definitely and  the  thrust  E  would  be  found  from  the  formula, 
Art.  44,  to  be,  for  w  =  58,  h  =  8, 

1  (cos*)* 

—  w  h2-     — —  =  1440  Ibs. 

2  cos  <p 

The  latter  thrust  acts  y£  AB  above  B  and  corresponds  to 
a  stress  uniformly  increasing  from  zero  at  A  to  360  lb./ft.2  at 
B,  "whereas  the  true  thrust  E  =  1150  Ibs.,  is  not  uniformly  in- 
creasing and  acts  about  0.35  to  0.36  AB  above  B,  Art.  43.  It 
will  probably  answer  the  purposes  of  computation  to  regard  the 
resultant  as  acting  at  the  third  point  and  the  stress  on  AB  as 
uniformly  increasing,  from  zero  at  A  to  1150  -5-  4  =  287  lb./ft.2 
at  B. 

*  In  the  case  of  a  very  acute  intersection  as  at  Q',  the  general  graphical 
method  of  Art.  32  gives  a  closer  result. 


224         THEORY   PERTAINING    TO   DEEP   AND    SHALLOW   BINS 

The  method  to  follow  in  the  case  of  the  vertical  side  of  the 
bin  AB  is  clear  and  precise.  The  same  cannot  be  said  for 
the  inclined  side  BC,  either  as  to  estimating  the  coal  thrust  on 
it  or  its  point  of  application.  Some  assumption  will  have  to  be 
made  and  the  simplest,  for  a  trial  solution,  is  to  assume  the 
thrust  on  the  vertical  plane  CN  as  acting  horizontally,  so  that 
V  =  o  for  this  plane. 

Hence,  having  drawn  CD  at  the  natural  slope,  to  intersection 
D  with  KH  produced,  draw  NM  II  CK  to  intersection  M  with 
EK  produced;  then  draw  MO  making  the  angle  <p  +  <?'  = 
v?  =  30°  with  NC  or  with  the  vertical  (whence  MO  A.  CD)  to 
intersection  0  with  CD.  With  the  mid-point  of  OD  as  center, 
describe  the  semi-circle  through  0  and  D  and  lay  off  CI  =  length 
of  tangent  to  semi-circle  from  C.  Then  draw  from  7  a  line  par- 
allel to  MO  to  intersection  Q  with  KH  and  lay  off  IL  =  IQ. 
It  follows  from  Art.  39,  that  CQ  is  the  plane  of  rupture  and 
that  the  thrust  is, 

E0  =  y2QPXw  =  y2  (i5-23)2  X  58  =  6730  Ibs. 

This  thrust  will  now  be  combined  with  the  weight  of  coal 
ABCN  =  W=  10,760  Ibs.  and  the  reaction  E  of  the  side  AB 
of  the  bin,  to  find  the  thrust  on  B  C.  Thus,  lay  off  to  scale, 
Ne  =  W,  ef  equal  and  parallel  to  E,  and  fh  equal  and  parallel 
to  E0\  whence  Nh  gives  the  amount  and  direction  of  the  thrust 
on  B  C.  In  magnitude  it  is  1 1 ,900  Ibs.  and  it  is  found  to  make 
an  angle  of  18°  with  the  normal  to  B  C. 

This  angle  is  very  near  the  maximum  allowable,  <pf  =  15°,  so 
that  the  thrust  is  perhaps  near  the  correct  value.  Referring  now 
to  the  general  conclusions  of  Art.  30  and  Art.  32,  since  BC  lies 
below  the  "  limiting  plane"  and  the  construction  above  leads 
to  a  thrust  on  BC  that  makes  a  greater  ansde  than  <?'  with  its 
normal,  it  is  next  in  order  to  assume  E\,  the  reaction  of  BC, 
as  making  the  angle  <pf  =  15°  with  the  normal  to  B  C  and  by 
a  construction  similar  to  that  of  Fig.  n,  Art.  32,  except  that 
the  reaction  E  is  introduced,  ascertain  the  magnitude  of  EI. 

This  second  trial  solution*  is  effected  by  drawing  trial  planes 

*  The  construction  is  so  similar  to  that  in  Fig.  77  or  in  Fig.  78,  with  the 
reactions  of  Gd,  Ge2,  .  .  .  ,  omitted,  that  it  is  not  given  here. 


110]  HOPPER  BIN.      COAL  HEAPED  225 

of  rupture  from  C  to  points  on  KH,  as  Q,  and  regarding  ABCQK, 
etc.,  as  trial  prisms  of  rupture,  each  subjected  to  four  forces  in 
equilibrium:  its  weight,  the  reaction  E,  the  reaction  Ei,  and 
the  reaction  of  the  trial  plane  of  rupture  making  the  angle  <p 
with  its  normal. 

The  weight  is  laid  off  vertically  downwards  to  scale,  say  from 
N  to  ei  (not  shown),  then  ej\  is  drawn  parallel  and  equal  to  E, 
then  a  line  from  /i  parallel  to  assumed  direction  of  EI  to  inter- 
section ci  with  line  from  N  parallel  to  the  reaction  of  trial  plane 
of  rupture.  Effecting  similar  constructions  for  the  various  trial 
prisms  of  rupture  and  reasoning  as  in  Art.  21,  it  is  seen  that  the 
greatest  length  of  the  type  fc,  gives  to  scale,  the  magnitude  of 
the  thrust  on  B  C,  EI  =  11,400  Ibs.,  which  by  assumption  is 
inclined  15°  to  the  normal  to  B  C.  It  is  found  also,  that  the 
corresponding  thrust  on  NC  is  inclined,  having  a  horizontal 
component,  acting  to  left,  =  7300  Ibs.  and  a  vertical  component 
acting  upward  =  700  pounds. 

The  result  strictly  refers  to  coal  not  restrained  by  the  walls, 
FG,  GH,  on  the  right.  Thus  if  we  suppose  C  F  =  o,  so  that 
the  bin  takes  the  shape  Fig.  73,  with  the  surface  as  per  dotted 
lines,  the  construction  just  indicated  gives  a  thrust  on  NC  more 
inclined  than  before,  whereas  the  restraint  of  the  bin  requires 
it  to  be  horizontal.  On  that  account,  the  result,  although 
perhaps,  a  fair  approximation,  is  not  entirely  satisfactory  and 
since  the  construction  demands  a  safe  value,  the  following 
approximate  construction  is  suggested: 

After  finding  E0  and  E  by  the  graphical  method  first  indicated 
construct  the  force  polygon  Nefh.  If  it  is  found  that  Nh 
makes  an  angle  with  the  normal  to  B  C  which  does  not  exceed 
<p'j  then  N  h  can  be  regarded  as  the  thrust  on  B  C.  If,  however, 
this  angle  exceeds  ^',  then  N I  is  drawn  making  the  angle  <pf  with 
the  normal  to  B  C  to  intersection  /  with  fh  extended  and  N I 
(to  scale)  is  to  be  taken  as  the  thrust  on  B  C.  For  the  bin, 
Fig.  74,  Nl  (to  scale)  =  EI  =  12,300  Ibs. 

Strictly,  the  stress  on  BC  is  not  uniformly  varying,  since 
the  surface  of  the  coal  AKH  is  not  a  uniform  slope,  and  the 
planes  of  rupture  at  different  depths  are  not  parallel.  The  law 


226        THEORY  PERTAINING  TO  DEEP  AND   SHALLOW  BINS 

of  uniform  variation  will,  however,  be  assumed  to  derive  for- 
mulas for  the  unit  stresses  at  B  and  C,  though  it  is  admittedly 
approximate. 

If  NA  and  C  B  are  extended  to  meet  at  n  and  any  line  is 
drawn  through  n,  cutting  perpendiculars  to  B  C  at  B  and  C, 
then  the  lengths  of  these  perpendiculars,  thus  limited  by  the 
line,  can  represent  the  magnitudes  of  the  (oblique)  unit  stresses 
qi  and  q  at  B  and  C;  also  the  ordinate  at  any  point  of  B  C  to  the 
line  will  represent  the  magnitude  of  the  stress  there,  according 
to  the  assumed  law  of  uniform  variation  and  the  area  of  the 
trapezoid,  having  q  and  q\  for  parallel  sides,  is  equal  to  E\  in 
magnitude.* 

For  brevity,  let  n  C  =  I,  n  B  =  a. 


K  (q  +  qi)  (I  —  &) ;  qi  =  q— 


On  substituting  the  value  of  qi  from  the  second  equation  in 
the  first  and  solving  for  <?,  we  obtain, 

q  =  2  E,  —   -. 

/2  —  a2 

By  measurement  on  the  drawing,  it  is  found  that  /  =  23.0', 
a  =  7.1'  and  since  E\  =  12,300  Ibs.,  .".  q  =  1180  lb./ft2,  q1  = 

1180  X  —  =  370  lb./ft.2 
23.0 

These  unit  stresses  at  C  and  B  act  parallel  to  N  I.  The 
position  of  the  resultant  of  all  the  stresses  on  BC  is  not  altered 
if  they  are  all  supposed  to  act  at  right  angles  to  B  C;  hence 
the  distance  y  from  C  to  where  the  resultant  EI  cuts  B  C,  will 
be  given  by  the  formula  of  Art.  34,  for  the  distance  to  the  center 
of  gravity  of  the  trapezoid  of  which  the  two  parallel  sides  are 
q  and  qi,  both  supposed  perpendicular  to  B  C.  We  have, 

hq+2q!      15.91180  +  2X370 

y  = = =  o.o  it. 

3     g  +  qi          3          Il8°  +  370 

*  An  error  that  is  sometimes  made,  is  to  lay  off  q  and  qi  parallel  to  EI  and 
to  regard  the  area  of  the  trapezoid  formed  with  q  and  qi  as  parallel  sides,  as 
the  magnitude  of  EI. 


1101 


HOPPER    BIN.      COAL   HEAPED 


227 


The  pockets  being  17  ft.  long,  the  interior,  frames,  shown  in 
Fig.  75,  are  17. ft.  apart. 

The  total  thrust  on  the  vertical  wall,  AB,  Fig.  74,  for  the 
17  ft.  length  is  17  X  1150  =  19,550  Ibs.;  hence  assuming  yz 
of  this  carried  to  the  framing  at  A  and  the  remainder  to  B 


FIG.  75 

have  the  loads  6500  and  13,000  Ibs.  at  A  and  B,  both  making 
ic  angle  $  =  15°  with  the  normal  to  AB,  to  be  used  in  estimat- 
the  stresses  in  the  framing.. 

Similarly,  the  total  thrust  for  17  ft.  length  of  face  B  C  is  17  X 
=  17  X  12,300  =  209,100  Ibs.,  whence  the  loads  at  B  and  C 
of  the  framing  are, 

6.6 

209,100  X =    86,735  Ibs., 

15-9 

9.3 

209,100  X =  122,265  Ibs., 

15-9 

both  loads  making  the  angle  <pf  =  15°  with  the  normal  to  B  C. 

Finally,  the  weight  of  coal  vertically  over  the  bottom  CF, 

for  17  ft.  length  of  bin  is  2  X  134,500  Ibs.,  so  that  the  loads 

corresponding  carried  to  one  frame,  at  C  and  F,  are  134,500 


228        THEORY  PERTAINING   TO  DEEP  AND   SHALLOW  BINS 

Ibs.  each.  The  total  weight  of  coal  carried  in  one  pocket  is 
2  X  318,000  Ibs.  and  this  should  equal  the  sum  of  the  vertical 
components  of  the  loads  at  all  the  joints  of  the  frame  shown  in 
Fig.  75.  From  these  loads,  the  stresses  in  the  framing  can  be 
ascertained,  either  by  graphics  or  by  taking  moments. 

Remark.  A  solution  that  has  been  proposed  for  this  type 
of  bin  is  as  follows:  extend  B  C  and  G  F,  Fig.  74,  to  intersection 
Kf  (not  shown)  and  find  by  the  construction,  the  horizontal  thrust 
on  KK'  to  be  8780  Ibs.  ;  combine  this  with  the  weight  of  n  KKf  to 
find  the  pressure  on  n  K'\  whence  the  unit  pressure  on  n  Kf  at 
Kf  can  be  computed,  assuming  the  stress  to  vary  uniformly  from 
zero  at  n\  whence  the  stresses  q  and  <?i  at  C  and  B  can  be  found 
and  the  position  and  amount  of  Ei  determined  as  above  ex- 
plained.* 

Assuming  the  thrusts  on  CN  and  BA  to  act  horizontally,  the 
true  thrusts  on  those  planes,  by  the  construction  of  Fig.  74,  are 
found  to  be  6728  and  1225  Ibs.  respectively.  But  corresponding 
to  the  law  of  uniformly  varying  stress  on  n  K',  the  thrusts  on 
the  vertical  planes  KK'  ,  NC  and  AB  must  vary  as  the  squares 
of  the  depths,  which  would  give  the  thrust  on  NC  =  8780 


NC2 

X  -  —  =  5090  Ibs.  and  the  thrust  on  AB  =  8780  X  —  —  = 
KK  2  KK 

487  Ibs.,  which  are  considerably  ±ess  than  the  above 

The  proposed  solution  considerably  underestimates  the  thrust 
on  NC  and  should  be  rejected.  A  closer  approximation  is  ob- 
tained by  assuming  the  thrust  on  NC  to  be  given  by  the  construc- 
tion, Fig.  74,  its  amount  being  6728  Ibs.,  whence  the  assumed 
horizontal  thrust  on  AB,  by  the  law  of  squares  above,  is  6728 

AB2 

-  =  645  Ibs.,  which  is  about  Jialf  the  true  thrust  on  AB, 

i\  (_/ 

corresponding  to  <pf  =  o.     The  influence  of  this  thrust  on  the 
equilibrium  polygon  Nefh  is,  however,  small. 

The  force  polygon  is  now  drawn  by  laying  off  Ne  =  W  = 
weight  of  coal  ABCN,  then  ef  =  645  Ibs.,  horizontally  to  right, 

*  Ketchum's  "Walls,  Bins,  and  Grain  Elevators,"  p.  192. 


110,  111] 


HOPPER  BIN.      COAL  HEAPED 


229 


fh  =  6728  Ibs.,  horizontally  to  left,  whence  Nh  =  12,400  Ibs.  = 
thrust  on  BC,  which  is  practically  the  thrust,  Nl  =  12,300  Ibs., 
found  above.  It  makes  an  angle  of  about  17°  with  the  normal 
to  B  C.  The  values  of  q,  qi  and  y  can  be  found  by  the  equations 
above.  Otherwise  the  thrust  on  NC,  6728  Ibs.,  can  be  com- 
bined with  the  weight  of  coal  n  NC  =  11,960  Ibs.,  supposed  to 
fill  this  space,  to  find  the  resultant  on  nC  =  13,700  Ibs.  .'. 
l/2  q  .nC  =  }4  q  X  23.1  =  13,700  .'.  q  =  1190  lb./ft.2  and 


=  1190  X  —  =  371  lb./ft.2 
23.1 


Whence  the  thrust  on  B  C  = 


^2  (q  +  qi)  X  B  C  =  780.5  X  15.9  =  12,410  Ibs.,  agreeing  with 
the  previous  determination. 

It  is  instructive  to  compare  the  results  of  the  four  approxi- 
mate solutions  given  above,  and  they  are  grouped  together  in 
the  following  table: 


Angle  Ei 

THRUST 

ON  NC 

Number 

Ei 

Normal  to 
BC 

Horl. 
Compt. 

Vert. 
Compt. 

I 

Lbs. 

11,900 

18° 

Lbs. 
6.71O 

Lbs. 
o 

2 

1  1  ,400 

IB 

7.7OO 

7OO 

7 

12,300 

1C 

7  COO 

o 

4  .  . 

12,400 

17 

6.71O 

o 

In  (i),  the  thrust  on  NC  is  assumed  to  act  horizontally.  It 
is  computed  for  the  actual  free  surface  and  combined  with  E 
and  W  to  find  EI  in  direction  and  magnitude. 

In  (2),  the  obliquity  of  EI  was  assumed  at  <p  =  15°  and 
its  magnitude  found  by  a  graphical  method. 

In  (3),  the  horizontal  thrust  on  NC  was  arbitrarily  increased, 
so  that  EI  should  make  the  angle  <p  =  15°  with  the  normal 
to  BC. 

The  solution  (4)  is  given  under  "  Remark,"  Art.  no. 

in.  Hopper  Bin.  Coal  Level  at  Top.  The  same  bin  and 
data,  only  with  the  coal  level  with  the  top,  will  be  treated, 
(i)  assuming,  as  before,  that  the  thrust  on  the  vertical  plane  Cg 


230        THEORY  PERTAINING   TO   DEEP   AND   SHALLOW   BINS 

is  horizontal,  Fig.  76.  By  the  construction  of  Art.  40,  we  find 
E,  the  thrust  per  linear  foot  on  wall  AB,  to  be  560  Ibs.,  the 
thrust  being  inclined  </  =  15°  to  the  horizontal.  Similarly,  the 
horizontal  thrust  on  the  vertical  plane  Cg,  is  found  to  be  E^  = 
3700  Ibs.  and  the  weight  of  coal  AB  Cg  =  W  =  8773  Ibs. 


FIG.  76. 


On  laying  off,  to  scale  vertically,  gk  =  W,  kl  equal  and 
parallel  to  E,  then  horizontally  to  left  from  /,  Ih  =  E0,  then 
gh,  the  resultant  on  B  C,  measures  9180  Ibs.  and  it  is  found  to 
make  an  angle  of  26°  with  the  normal  to  B  C,  which  exceeds 
the  maximum  obliquity  allowable,  <pr  =  15°,  by  11°. 

(2)  By  the  reasoning  of  Art.  29,  it  is  next  in  order  to  assume 
that  the  unknown  thrust  on  BC  makes  an  angle  of  15°  with 
the  normal  to  B  C  and  by  a  construction  similar  to  the  "  second 
trial  solution"  of  Art.  no,  to  find  the  magnitude  of  this  thrust. 
It  is  found  to  be  8100  Ibs.  The  construction  likewise  shows 
that  the  active  thrust  on  the  vertical  plane  Cg  has  a  horizontal 
component  of  4800  Ibs.  and  a  vertical,  upward  component  of 
1700  Ibs.;  also  that  the  result  would  be  the  sar^e  if  CD,  Fig.  76, 
was  zero  or  for  the  bin  of  Fig.  73.  But  in  the  latter  figure,  the 
thrust  on  the  medial  plane,  CN,  is  horizontal  and  not  consider- 
ably inclined  as  above.  The  method  then  fails  for  this  bin, 
Fig.  73,  which  must  be  attributed  to  the  restraining  influence 
of  the  walls  in  the  right  half  of  the  bin  and  this  throws  doubt, 
though  to  a  less  extent,  upon  the  construction  to  evaluate  Ei 
for  the  bin,  Fig.  76.  Jt  may  be  seen  also,  that  the  upward 


Ill] 


HOPPER  BIN.   COAL  LEVEL  AT  TOP 


231 


acting  vertical  component  on  Cg  diminishes  the  value  of  EI 
materially  as  found  by  the  first  trial  solution. 

(3)  As  the  case  does  not  seem  to  admit  of  an  exact  solution, 
the  following  procedure  is  suggested:    draw  the  line  gn  making 
the  angle  <?'  —  15°  with  the  normal  to  B  C  to  intersection  n 
with  the  horizontal  through  I  and  regard  gn  as  the  thrust  on 
B  C.    Its  magnitude  is  EI  =  10,100  Ibs.     This  is  decidedly  on 
the  side  of  safety,  since  the  horizontal  component  In  —  5800 
Ibs.  is  greater  than  in  either  trial  solution,  so  that  a  certain 
amount  of  passive  thrust  is  involved. 

(4)  Turning  now  to  the  last  solution  indicated  in  "  Remark/' 
Art.  no  and  regarding  AB  as  a  perfectly  smooth  wall,  we  have 
strictly,  for  the  plane  free  surface  AG,  E/E0  =  AB*/Cg*,  so 
that  the  stress  on  B  C  is  uniformly  increasing.     In  fact,  if  the 
wall  AB  is  removed,  a  wall  Ba  added  and  the  space  B  a  A 
supposed  filled  with  coal,  the  horizontal  thrust  on  the  vertical 
plane  AB  is  the  same  as  before  and  the  magnitude  of  the  stress 
on   a  C   increases   uniformly  from  zero  at  a  to  a  maximum 
Cc  at  C. 

On  combining  E0  with  weight  of  coal  a  Cg  it  is  found  that 
the  resultant  on  aC,  makes  an  angle  26°+  with  the  normal 
to  a  C.  The  magnitude  of  this  resultant  can  be  represented  by 
the  area  of  the  right  triangle  a  Cc',  whence  unit  stress  at  C  = 
resultant  divided  by  %  a  C,  whence  the  unit  stress  at  B  can  be 
found  and  area  trapezoid  Bb  Cc  =  magnitude  of  EI  =  9380  Ibs. 
By  the  formula  of  Art.  no,  it  cuts  BC  at  the  distance  y  above 
C  and  its  direction  is  that  of  the  resultant  on  a  C. 

The  results  of  the  four  approximate  solutions  are  recorded 
in  the  following  table: 


Angle  Made 

THRUST 

ON  C?, 

Number 

Ei 

by  .til 
with  Normal 
toBC 

Horl. 
Compt. 

Vert. 
Compt. 

Lbs. 

Q  1  80 

26° 

Lbs. 

37OO 

Lbs. 

Q 

2  

8  100 

T  e 

,/uu 
4  8(X) 

I     7OO 

3  

IO  IOO 

1D 
T  e 

58OO 

Q 

4  

Q  ^80 

26 

37OO 

Q 

232         THEORY   PERTAINING   TO   DEEP   AND    SHALLOW   BINS 

The  values  of  Ei  differ  considerably — the  true  value  evidently 
lying  between  the  extremes.  For  the  value  of  EI  selected,  say 
No.  3,  the  quantities  q,  qi  and  y  can  be  found  from  the  formulas 
of  Art.  no  and  the  solution  completed  exactly  as  in  that  article. 

Presumably  the  bin  will  be  designed  for  the  "coal  heaped," 
Fig.  74,  where  the  agreement  between  the  values  for  E\  is  much 
more  satisfactory  than  for  the  case  just  treated. 

112.  Unsymmetrical  Bin.  Let  us  now  consider  the  coal  bin 
ABCDFG,  Fig.  77,  which  only  differs  from  the  bin  of  Fig.  74 
in  the  prolongation  of  the  vertical  wall  FG  to  meet  the  surface 
of  the  coal,  sloping  at  the  angle  of  repose,  <p  =  30°  from  A 
upwards  to  the  right. 

As  before,  assume  w  —  58  lb./ft.3,  </  =  15°,  so  that  the 
pressure  on  any  side  cannot  make  an  angle  with  its  normal 


FIG.  77. 


that  exceeds  the  angle  of  friction  of  coal  on  steel,  <?'  =  15°. 
The  value  of  the  thrust  £4  on  FG  is  quickly  found,  by  the  con- 
struction of  Fig.  20,  Art.  40,  to  equal  4550  Ibs.  It  ac.ts  Y$  FG 
above  F.  To  find  the  thrust  £3  on  the  wall  DFi'by  Art.  30, 
the  first  step  would  be  to  find  the  thrust  on  the  vertical  plane 
DO  and  then  combine  it  with  £4  and  TFi,  the  weight  of  coal 
ODFG,  to  find  the  thrust  E3  in  amount  and  direction.  This 


112]  UNSYMMETRICAL  BIN  233 

would  be  the  true  thrust  on  DF;  unless  its  obliquity  exceeded 
<pf,  in  which  case,  the  obliquity  of  £3  is  assumed  equal  to  <pf 
and  the  magnitude  of  Es  determined  by  the  graphical  construc- 
tion given  below.  Unfortunately  the  direction  of  the  thrust 
on  D  0  is  not  known  so  that  its  magnitude  cannot  be  found. 
In  fact,  when  this  direction  is  near  that  of  A  G,  as  in  this  case, 
a  few  degrees'  change  in  this  direction  will  make  a  large  alteration 
in  the  magnitude  of  the  thrust.  Hence  in  this  case,  the  first 
test  cannot  be  applied  and  the  second  is  alone  available. 

By  this  method,  trial  planes  of  rupture  Dbi,  Dbz,  Dbs,  .  .  ., 
are  assumed  and  the  weights  of  the  prisms  GFDO,  GFDbi, 
GFDbz,  .  .  .,  computed  and  laid  off,  to  scale,  say  from  g,  verti- 
cally downward  to  o,  i,  2,  etc.  At  these  points,  draw  lines  to 
the  left  equal  and  parallel  to  £4,  to  points  such  as  /  and  n. 
From  such  points,  draw  lines  parallel  to  E3  (assumed  to  make 
an  angle  of  <?'  =  15°  with  the  normal  to  DF)  to  intersections 
CQ,  Cij  Cz,  •  •  .,  with  the  rays  gs,  gsi,  gSz,  .  .  .,  which  make  the 
angles  y  below  the  normals  to  planes  DO,  Dbi,  Dbz,  .  .  .  .  To 
construct  these  rays,  with  D  and  g  as  centers  and  a  convenient 
radius  =  DaQ  =  gs,  describe  arcs  of  circles  #o  #3  and  s  $3. 
Let  DO,  Dbi,  Dbz,  .  .  .,  cut  the  first  arc  at  a0,  fli,  #2,  •  •  ., 
and  suppose  gs  drawn  parallel  to  AG  so  that  it  makes  the  angle 
<p  with  the  normal  to  the  vertical  plane  D  0.  Then  if  we  lay 
off  s  Si  =  aQ  ai,  s  s2  —  a0  a2,  •  •  .,  the  lines  gsi,  gs2,  .  .  '.,  evi- 
dently make  the  angle  <p  with  the  normals  to  planes  Dbi,  Db2, 
.  .  .,  and  thus  give  the  required  directions  of  the  reactions  of 
those  planes — each  taken  in  turn  as  a  possible  plane  of  rupture. 

On  drawing  a  vertical  tangent  to  the  curve  through  CQ,  c\, .  .  ., 
which  happens  to  pass  through  cs,  the  true  thrust  E3  is  given 
by  the  line  nc$  and  the  true  (and  only)  plane  of  rupture  is  Abz. 
The  proof  is  as  follows:  the  prism  G F  D  bs  is  in  equilibrium 
under  the  action  of  four  forces:  its  weight,  E4,  E3  and  the  re- 
action of  Abs',  represented  by  #3,  $n,  nc^  and  csg,  the  sides  of 
a  closed  polygon.  The  line  ncs  is  the  greatest  of  the  lines  drawn 
parallel  to  jE3  and  represents  its  magnitude;  since  n  being 
fixed,  if  ncz  is  shortened,  cz  falls  below  its  present  position,  so 
that  the  resulting  reaction  c3g  makes  an  angle  greater  than  <f> 


234        THEORY   PERTAINING  TO  DEEP  AND   SHALLOW  BINS 

with  the  normal  to  the  plane,  Ab$,  which  is  inconsistent  with 
equilibrium.  Measuring  ncs  to  the  scale  of  loads,  we  find 
Es  =  14,300  pounds. 

On  producing  fc0  to  the  vertical  tangent  to  d,  so  that  fd  = 
nc$,  it  is  seen  that  dg  represents  the  thrust  on  the  vertical  plane 
D  O,  since  GFDO  is  in  equilibrium  under  its  weight,  the  reactions 
£4  and  Es  and  the  thrust  on  DO  acting  to  the  right,  and  these 
forces  are  proportional  to  the  sides  of  the  closed  polygon  gofdg. 
On  measuring  dg  to  the  scale  of  loads  it  is  found  that,  dg  = 
thrust  on  plane  D  O  =  12,700  Ibs. 

As  a  check,  if  the  direction  of  the  thrust  on  D  0  is  assumed 
to  be  parallel  to  dg  and  the  successive  trial  prisms  of  rupture 
are  ODbi,  ODb%,  .  .  .,  then  either  the  general  graphical  method 
of  Art.  21  or  that  of  Art.  40,  leads  to  the  same  thrust  on  DO 
of  12,700  Ibs. 

If  the  thrust  on  DO  is  assumed  to  act  parallel  to  AG,  its 
magnitude,  for  an  unlimited  mass  of  coal,  would  be,  Art.  46, 

y£  w.  D  O2  cos  <p  =  y£  (58)  (31. 5)2  cos  30°  =  24,900  Ibs., 
which  is  nearly  twice  the  preceding  value.     The  result  is  in- 
applicable to  this  problem,  since  it  involves  an  indefinitely  great 
prism  of  rupture  on  the  right  of  D  0,  Art.  41,  whereas  the  walls 
of  the  bin  limit  this  prism. 

The  magnitude  of  £3  having  been  determined,  its  point  of 
application  on  DF  can  be  found  as  in  Art.  1 10,  the  point  of  inter- 
section of  DF  and  AG  produced,  corresponding  to  the  point  n 
of  Fig.  74.  Practically,  for  this  bin,  £3  may  be  taken  as  acting 
at  the  mid-point  of  DF. 

Thrust  on  B  C.  In  Fig.  78  is  given  the  construction  for 
finding  the  thrust  EI  on  B  C.  The  possible  prisms  of  rupture 
are  ABC G,  ABCe^G,  ABCe2G,  .  .  .,  and  their  weights  are 
laid  off  from  a  convenient  point  F,  vertically  downwards,  to 
o,  i,  2,  .  .  .,  respectively.  By  the  construction  of  Art.  40, 
we  find  E  =  1440  Ibs.  It  acts  %  AB  above  B. 

At  the  points  o,  i,  2,  .  .  .,  draw  lines  equal  and  parallel  to 
E  (which  is  inclined  15°  to  the  horizontal)  to  the  right,  to  locate 
points  such  as/  b. 

It  is  necessary  now  to  compute  the  reactions,  acting  parallel 


112] 


UNSYMMETRICAL   BIN 


235 


to  £4,  of  the  portions  of  the  wall,  Ge±,  Ge2,  .  .  .,  which  is  easily 
done,  since  the  thrust  on  FG  =  E4  =  4550  Ibs.  and  the  thrust 
varies  as  the  square  of  the  depth.  Thus  the  thrusts  on  Gei, 

Get       Get 
Ge2,.  .  .,are  — E4,— ;£*,    .  ,  ,    where       FG  ==  26.35   ft., 

Gei  =  5  ft.,  Gc2  =  10  ft.,  etc. 

These  thrusts  onGei,  Ge2,  .  .  .,  are  to  be  laid  off  to  the  left, 
parallel  to  E^  from  the  extremities  of  the  lines  drawn  represent- 


^"'^'"'',' 

n-fi;  / 


FIG.  78. 

ing  E.    Thus  at  /,  jg  is  drawn  parallel  to  E4  and  equal  to  the 
reaction  of  Ge^  etc. 

From  the  extremities  of  the  last  lines  drawn,  or  from  points 
such  as  g,  lines  are  drawn  parallel  to  the  assumed  direction  of 
EI  (which  makes  an  angle  of  15°  with  the  normal  to  B  C)  to 
intersections  CQ,  Ci,  c^  .  .  .,  with  the  rays  from  V  inclined  v>° 
to  the  normals  to  planes  CG,  Ce\,  Cez,  ....  The  directions  of 
these  rays  are  easily  found,  Art.  19,  by  first  drawing  C  d  making 


236        THEORY  PERTAINING  TO  DEEP  AND   SHALLOW  BINS 

the  angle  <p  =  30°  with  the  horizontal;  then  with  C  and  V  as 
centers  and  with  the  same  radius,  Vs  =  C  d,  describing  arcs 
dao,  SSQ,  and  then  laying  off  SSQ  =  da0,  ssi  =  daij  .  .  .,  whence 
Vs0,  Vsi,  .  .  .,  make  angles  <p  with  the  normals  to  planes  CG, 
Ceij  .  .  .,  as  required. 

The  lines  representing  possible  values  of  EI,  are  those  drawn 
parallel  to  E\  and  terminating  in  CQ,  Ci,  c%,  c$.  Of  these  lines, 
gcz  is  the  greatest  and  represents  the  true  thrust  Ei.  This  is 
plain,  if  we  note  that  V2/g  c2  V  is  a  closed  polygon,  whose  sides 
represent  in  turn,  the  weight  of  prism  ABCe2G,  the  reaction 
E  of  wall  AB,  the  reaction  (acting  parallel  to  E±)  of  part  of 
wall  Ge2,  the  reaction  EI  of  wall  B  C  and  lastly,  the  reaction 
CzV  of  plane  C  e2.  Similar  closed  polygons  refer  to  the  other  trial 
prisms  of  rupture.  Now  as  g  is  fixed,  the  reaction  gc2  cannot  be 
diminished  except  by  lowering  c2;  but  this  would  cause  czV 
to  make  a  greater  angle  than  <p  with  the  normal  to  plane  Abz, 
which  is  inconsistent  with  equilibrium. 

The  line  g  cz  thus  represents  EI.  Its  value,  to  the  scale  of 
loads,  is  EI  =  14,600  pounds.  Its  position  on  B  C  is  found  by 
using  the  formulas  of  Art.  no. 

The  next  step  is  to  find  the  thrust  on  the  vertical  plane  NC. 
To  do  this,  lay  off  Vn  vertically  downwards,  equal  to  weight  of 
prism  AB  CN',  then,  n  i  equal  and  parallel  to  E\  next,  i  h  equal 
and  parallel  to  gc2,  representing  Ei;  whence  hV  represents  the 
coal  thrust  on  CN  in  amount  and  direction.  Its  value  is 
9200  pounds. 

It  was  ascertained  above  that  the  thrust  on  plane  D  0  was 
12,800  Ibs.  Lay  this  off  from  h,  Fig.  78,  to  the  scale  of  loads 
used  in  this  figure,  parallel  to  its  position  gd  given  in  Fig.  77, 
to  point  r  and  from  r,  lay  off  vertically  dowr.  ^ards,  rt  =  weight 
of  coal  in  prism  NCDO  =  15,600  Ibs.  Then  Vt  (to  scale)  = 
18,700  Ibs.  =  E2  =  pressure  on  bottom  CD,  acting  parallel  to 
Vt.  This  follows,  because  the  prism  of  coal  NCDO,  is  subjected 
to  the  thrust  on  N  C  =  Vh,  directed  from  left  to  right,  a  thrust 
on  D  0  =  hr,  directed  from  right  to  left,  its  weight  =  rt  and 
the  reaction  E2  of  CD,  which  must  equal  tV  to  close  the  polygon 
VhrtV,  the  sides  of  which  represent  the  four  forces  in  equilib- 


112]  UNSYMMETRICAL  BIN  237 

num.  The  point  where  E%  cuts  CD  is  uncertain — possibly  it  is 
near  the  center  of  CD. 

The  five  reactions  E,  EL,  £2,  E^,  E4,  have  now  been  com- 
puted. They  refer  to  i  ft.  length  of  bin.  If  they  are  plotted 
in  order,  to  scale,  it  will  be  found  that  a  vertical  line  is  necessary 
to  close  the  polygon  of  length,  to  scale,  equal  to  the  total  weight 
of  coal,  45,800  Ibs.,  thus  checking  the  work.  Of  the  above 
five  reactions,  E  and  £4  are  correct  and  E2  depends  to  a  small 
extent  on  the  thrusts  on  the  vertical  planes  CN  and  D  O,  which 
are  in  turn  connected  with  the  reactions  on  B  C  and  DF\  hence 
the  degree  of  reliability  of  the  constructions  for  rinding  EI  and 
£3  is  of  practical  interest. 

With  regard  to  E\,  the  analysis  is  along  similar  lines  to  that 
for  the  bin,  Fig.  74,  for  the  coal  heaped,  in  the  "second  trial 
solution,'7  which  led  to  fairly  good  results.  We  should  expect 
no  less  in  the  determination  of  EI  in  Fig.  78,  with  the  resulting 
construction  for  Vh,  the  thrust  on  NC.  The  latter  thrust  is 
nearly  parallel  to  AG,  which  seems  reasonable. 

The  construction  for  Es,  led  to  a  thrust  on  the  vertical  plane 
DO  oi  12,700  Ibs.,  acting  parallel  to  rh,  Fig.  78.  It  may  be 
thought  that  this  thrust  should  be  more  nearly  parallel  to  Vh, 
but  it  will  be  found  that  such  a  supposition  leads  to  an  obliquity 
for  £3  greater  than  15°  which  is  not  admissible.  Thus,  if  the 
direction  of  the  thrust  on  D  O  is  assumed  parallel  to  Vh,  the 
value  of  the  thrust  is  found  (by  the  method  of  Art.  40)  to  be 
9400  Ibs.  When  this  is  combined  with  E4  and  the  weight  of 
0  D  F  G,  it  is  found  to  give  a  thrust  on  D  F,  6%  greater  than 
before  and  making  a  greater  angle  than  15°  with  the  normal 
to  D  F.  Therefore,  for  such  cases,  by  Art.  29,  the  construction 
of  Fig.  77  is  alone  admissible  and  this  leads  to  the  value  of  Es 
given.  From  the  various  considerations  above,  it  is  thought 
that  the  values  found  above  for  EI,  E2j  E$,  are  fairly  good 
ones  and  that  the  constructions  for  finding  them  should  prove 
serviceable  to  the  constructor. 


APPENDIX  I 
Stresses  in  Wedge-shaped  Reinforced  Concrete  Beams 

i.  It  is  often  necessary  to  find  the  stresses  in  reinforced 
concrete  beams  whose  faces  are  inclined  to  each  other,  as  in  the 
case  of  the  toes,  heels,  and  counterforts  of  reinforced  concrete 
retaining  walls.  Only  an  approximate  solution  can  be  offered 
for  the  general  case,  which,  however,  is  on  the  side  of  safety 
and  reduces  to  well-known  results  when  the  beam  is  taken  as 
prismatic. 

In  Fig.  i  (i)  is  shown  a  portion  of  a  toe,  with  an  upper  face 
making  the  angle  0  with  the  horizontal  and  subjected  to  a 


(2) 


(3) 


FIG. 


load  Q  on  the  lower  horizontal  face,  acting  vertically  upward. 
The  reinforcement,  near  and  parallel  to  the  lower  face,  is  in 
tension.  As  usual,  no  tension  will  be  supposed  to  exist  along 
the  vertical  section  IN,  between  the  steel  at  /  and  the  neutral 
axis  0.  On  the  portion  ON,  both  compressive  and  ^hearing 
stresses  are  exerted. 

The  resultant  unit  stress  at  N  is  a  principal  compressive 
stress,  acting  parallel  to  the  surface  on  a  plane  at  right  angles 
to  it;  for  consider  a  small  cube  of  concrete  at  N,  with  faces 

239 


240  APPENDIX  I 

parallel  and  perpendicular  to  the  upper  surface.  Since  there 
is  no  shear  on  this  surface  (there  being  no  matter  above  it  to 
produce  shear),  it  follows  from  a  well-known  theorem  that 
there  can  be  no  shear  on  planes  perpendicular  to  it;  conse- 
quently, the  unit  stress  p0  on  a  plane  at  N,  at  right  angles  to 
the  surface,  is  normal  to  it;  hence  p0  is  a  principal  compressive 
stress,  and  it  acts  parallel  to  the  upper  face. 

2.  It  will  next  be  shown  that  if  f0  is  the  horizontal  com- 
ponent of  the  unit  stress  at  N  on  the  plane  IN,  corresponding 
to  the  bending  moment  Qa,  then, 

p0  =  f0  sec*  ft. 

To  prove  this  important  formula,  consider  the  forces  keeping 
the  small  prism  ANB,  Fig.   2,  in  equilibrium;  the  horizontal 
component  of  the  unit  stress    on    the   vertical 
plane  AB  being  j0  and  the  unit  stress  on  the 
plane  AN,  perpendicular  to  the  upper  surface 
NB,  being  p0.    There  is  a  shear  on   AB,  but 
none  on  AN  or  NB.    When    AN    is    infini- 
FIG.  2          tesimal,   the  total  stress  on  it  is  p0  .  AN  =  p0  . 
A B  cos  ft,  its  horizontal  component  is  p0  .  AB 
cos*  ft,  and  for  equilibrium  of  ANB  this  must  equal  the  total 
horizontal  component  of  the  stress  on  AB  or/0 .  AB. 

On  equating  and  solving,  we  derive,  p0  =       °      =  f0  sec2  ft. 

cos2  ft 

3.  Let  us  suppose,  as  an  approximation,  that/0  is  computed 
by  the  ordinary  formula  for  a  prismatic  beam,  just  as  if  ft  was 
zero,  so  that  f0  remains  the  same  for  any  ft.     It  would  follow 
from   the  exact  formula  p0  =  f0  sec2  ft,  as  ft  approaches   90°, 
that  p0  would  become  indefinitely  large,  whicL  is  absurd.     The 
proposed  approximation  thus  fails  to  give  reasonable  results  for 
large  values  of  ft. 

When  ft  —  o,  p0  =  f0  is  an  accepted  value,  but  as  ft  increases, 
it  seems  essential  that/0  should  decrease  continually  so  that  p0 
shall  remain  within  reasonable  limits.  The  reason  is  evidently 
that  the  toe  to  the  left  of  IN,  Fig.  i  (i),  is  less  resisting  than  a 


3,  4]  ASSUMPTIONS  241 

prismatic  beam  of  the  same  length,  and  hence  f0  for  the  toe  is 
less  than  for  the  prismatic  beam,  and  the  conservation  of  plane 
sections  after  strain  may  not  apply  there. 

It  appears  further  that  the  excess  in  the  computed  value 
of  f0  for  the  prismatic  beam,  over  its  true  value  for  the  toe, 
should  be  very  small  near  the  left  or  truncated  end  of  the  toe, 
since  the  resistances,  for  prismatic  beam  and  toe,  as  regards 
bending,  are  now  nearly  equal,  the  common  section  IN  being 
now  near  the  end  of  the  toe.  As  the  section  IN  is  taken  farther 
from  the  end  of  the  toe,  the. excess  should  increase  and  reach  a 
maximum  where  the  toe  joins  the  stem. 

The  state  of  the  horizontal  stress  on  77V  is  then  not  uniformly 
increasing  from  0  to  AT"  as  indicated  in  Fig.  i  (i),  but  the  true 
unit  stress  at  N  is  less  than  as  computed  for  a  prismatic  beam, 
the  unit  stress  then  probably  increasing  for  a  little  distance 
below  N  and  then  decreasing  as  O  is  approached,  since  the  sum 
of  the  moments  of  all  the  stresses  on  the  section  IN  about  I 
must  equal  the  constant  external  moment  Qa,  whether  the  unit 
stresses  are  taken  as  uniformly  increasing  or  as  just  indicated. 
These  inferences  all  receive  confirmation  from  the  analysis  of 
the  experiments  on  india-rubber  dams  of  Wilson  and  Gore  and 
those  on  plasticine  dams  of  Ottley  and  Brightmore,*  a  short 
discussion,  pertaining  to  this  subject,  being  given  by  the  writer 
in  Trans.  Am.  Soc.  C.  E.,  vol.  LXXVIII  (1915),  p.  747. 

The  assumption  that/0  is  to  be  computed  as  for  a  prismatic 
beam  thus  leads  to  an  excess  over  the  true  stress  at  N,  this 
excess  being  small,  say  for  0  less  than  20°,  but  increasing  more 
rapidly  for  larger  values  of  /3,  so  that  the  assumption,  for  lack 
of  definite  experimental  data,  will  be  limited  arbitrarily  to 
values  of  0  less  than  45°. 

4.  From  analogy  to  the  figures  given  by  Morleyf  or  Tur- 
neaure  and  Maurer,t  an  attempt  has  been  made  to  sketch  in 
Fig.  i  (3)  the  principal  compressive  stresses  at  various  points 

*  Minutes  of  Proceedings  of  Inst.  C.  E.,  vol.  CLXXII,  session  1907-08, 
Part  ii,  PI.  3,  Fig.  4,  and  PI.  5,  Figs.  19-21. 
t  "Strength  of  Materials,"  p.  138. 
J  "Principles  of  Reinforced  Concrete  Construction,"  Art.  46. 


242  APPENDIX   I 

of  the  section  IN.  A  principal  stress  is  to  be  defined  as  a  stress 
which  acts  on  a  plane  at  right  angles  to  its  direction,  there 
being  no  shearing  stress  on  the  plane.  It  thus  acts  normally 
to  the  plane  and  is  the  whole  stress  on  it. 

Besides  the  principal  compre"ssive  stresses  shown  in  Fig.  i  (3), 
there  are  principal  tensile  stresses,  at  right  angles  to  the  former, 
not  drawn,  but  which  must  be  included  when  the  whole  state 
of  stress  on  ON  is  considered.* 

5.  Fig.  i  (2)  shows  a  second  approximate  distribution  of 
stress  over  -the  section  IN,  the  stresses  uniformly  increasing  from 
a  point  0  and  all  acting  parallel  to  the  surface.  In  addition, 
there  is  a  shearing  stress  on  IN,  which,  added  to  the  sum  of 
the  vertical  components  of  the  parallel  stresses,  must  exactly 
equal  Q  or  the  shear  due  to  the  loading  of  whatever  character. 

The  writer  was  led  to  make  this  distribution  of  stress  be- 
cause, as  shown  above,  the  stress  at  N  acts  parallel  to  the  sur- 
face, and  in  the  formulas  that  will  be  derived,  this  stress  and 
the  angle  |8  are  included  from  the  start,  so  that  the  results  prom- 
ised to  be  more  reliable  than  those  pertaining  to  the  state  of 
stress  Fig.  i  (i). 

Before  taking  up  the  derivation  of  the  formulas,  it  is  well  to 
show  the  relation  between  the  state  of  stress  assumed  in  Fig.  i 
(2)  and  the  true  state  represented  by  Fig.  i  (3),  when  the  tensile 
principal  stresses  acting  on  ON  are  added.  This  relation  can 
be  easily  shown  in  a  manner  kindly  given  by  letter  by  Prof. 
H.  A.  Thomas,t  involving  the  use  of  the  ellipse  of  stress.  Thus 
in  Fig.  3,  at  a  given  point  P  on  ON,  let  c  —  principal  compres- 
sive  unit  stress  and  /  =  principal  tensile  unit  stress  (at  right 
angles  to  c);  then  from  the  ellipse  of  stress,  drawn  with  c  and 
/  as  semi-axes,  the  resultant  unit  stress  r  on  the  plane  IN  at  P 


*  Prof.  O.  H.  Basquin  has  kindly  sent  the  writer  what  he  styles  a  "Rough 
Test  of  a' Wedge-Shaped  Glass  Beam,"  with  polariscope,  to  show  the  inclina- 
tion of  the  principal  compressive  stresses  to  the  horizontal.  It  is  interesting 
to  observe  from  the  results  that  for  about  one-fourth  of  the  way  down  this 
inclination  was  practically  constant  and  equal  to  ft  =  l5-5°  f°r  the  beam 
experimented  with. 

t  Rose  Polytechnic  Institute,  Terre  Haute,  Ind. 


5,6] 


USE    OF    THE   ELLIPSE    OF    STRESS 


243 


can  be  found.  Otherwise,  Rankine's  well-known  construction 
for  finding  r  can  be  used.  This  resultant  r  can  be  decomposed 
into  a  component  p  parallel  to  the  compressed  surface  and  a 
vertical  component  q. 

A   similar   construction   being   supposed,  effected   at   every 
point  of  ON,  the  fundamental  assumption  made  by  the  writer 


FIG.  3 


can  be  thus  stated:  the  component  of  the  resultant  stress  r, 
parallel  to  the  surface  in  compression,  at  any  point  of  the  section 
IN,  is  assumed  to  increase  directly  as  the  distance  of  the  point 
from  the  neutral  axis.  The  neutral  axis  may  be  denned  as  the 
line  where  the  parallel  component  vanishes. 

This  is  in  agreement  with  the  basis  for  the  common  theory 
of  flexure  for  prismatic  beams,  which  may  be  thus  stated:  the 
normal  component  of  the  resultant  stress  on  the  reference  section 
at  any  point  is  assumed  to  increase  directly  as  the  distance 
of  the  point  from  the  neutral  axis.  This  use  by  Prof.  Thomas 
of  the  ellipse  of  stress  furnishes  a  scientific  basis  for  the  system 
of  parallel  stresses  proposed. 

6.  It  is  seen  that  where  the  construction  is  supposed,  made 
for  each  point  of  the  section  from  0  to  N,  the  result  will  be  not 
only  the  system  of  parallel  stresses,  Fig.  i  (2),  but  in  addition 
a  vertical  shear,  acting  downward  along  NO,  equal  to  Zg,  or  to 
the  sum  of  the  vertical  components  q  above  referred  to.  The 
total  vertical  shear  on  ON  is  thus  2q  +  the  sum  of  the  vertical 
components  of  the  parallel  stresses. 

The  point  0  in  Fig.  i  (2)  is  to  be  determined  by  a  formula 


244  APPENDIX  I 

derived  further  on.  If  the  point  so  determined  is  identical 
with  the  true  neutral  point  O  of  Fig.  i  (3),  then  the  two  systems 
of  stresses  Figs,  i  (2)  and  (3)  [with  the  tensile  principal  stresses 
added]  are  equivalent  and  the  moment  of  either  system  about  7 
is  the  same. 

However,  as  noted  above,  the  parallel  stresses  cannot  be 
regarded  strictly  as  varying  uniformly  for  a  wedge-shaped 
beam,  so  that  the  determination  of  0  on  that  basis  can  be 
regarded  as  only  approximative — sufficiently  near  for  the  smaller 
values  of  /3,  but  departing  more  and  more  from  accuracy  as  0 
increases.  Until  direct  experiment  can  speak  definitely  on  this 
subject,  it  is  suggested  that  the  applications  of  the  theory  that 
follows  be  limited  to  angles  of  /3  not  exceeding  45°. 

The  stress  p  in  Fig.  3  is  an  oblique  stress  per  unit  of  area 
of  the  plane  IN.  Thus  if  we  consider  a  cylindrical  fiber,  parallel 
to  the  compressed  surface  whose  section  in  the  plane  IN  has  an 
area  Aa,  the  total  stress  on  the  area  is  p  Aa.  As  it  is  desirable 
to  refer  the  stress  to  a  plane  perpendicular  to  p  or  to  the  fiber, 
let  /  equal  the  normal  unit  stress  on  such  a  plane.  The  area  of 
a  section  of  the  fiber  at  right  angles  to  its  length  is  A  a  cos  0 
and  the  total  stress  on  it  is  /  A#  cos  |8.  Since  this  must  equal 
p  .  Aa,  we  find  the  relation  /  =  p  sec  0;  hence  /  varies  with  p 
and  thus  follows  the  linear  law  of  variation.  In  Fig.  i  (2), 
any  one  of  the  parallel  unit  stresses  may  be  designated  by  the 
letter  /  and  the  unit  stress  at  the  surface,  which  acts  on  a  plane 
perpendicular  to  it,  will  be  designated  by  the  letter/,..  As  shown 
above,  this  is  a  principal  stress. 

The  previous  discussion  refers  to  the  most  difficult  case: 
that  where  the  compressed  surface  is  not  at  right  angles  to  the 
section  JN,  which  is  parallel  to  the  loads.  If  in  Fig.  i,  the  load 
Q  acts  down,  then  the  reinforcement  must  be  placed  near  the 
upper  inclined  surface  and  the  lower  horizontal  surface  will  be 
in  compression.  By  the  same  assumption  of  parallel  stresses, 
the  compressive  stresses  on  10  will  be  supposed  to  act  normally 
to  IN  or  parallel  to  the  lower  surface,  as  in  the  common 
theory. 

7.  A  general  solution  will  now  be  given  which  includes  every 


7] 


GENERAL   SOLUTION 


245 


possible  case.    All  special  cases  can  be  at  once  derived  from 
this  general  solution 

The  usual  hypotheses,  that  no  tension  exists  in  the  concrete 
on  the  reinforced  side  of  the  neutral  axis  and  that  plane  sections, 
remain  such  after  stress,  will  be  adopted. 


(5) 

STRAINS 

FIG.  4 


CROSS-SECTION 


In  (a)  and  (6),  Fig.  4,  are  shown  two  longitudinal  sections, 
and  in  (c)  a  section  of  a  part  of  a  beam  subjected  to  stress. 
IN  represents  a  section  of  the  beam  taken  always  parallel  to 
the  direction  of  the  loads,  which  may  be  weights,  soil  reactions, 
earth  thrusts,  etc.  The  shear  due  to  the  loads  thus  acts  along 
IN,  and  the  moment  is  the  same  for  any  point  of  the  section. 

For  the  breadth,  b,  let  A\,  A^  .  .  .,  represent  the  areas  of 
the  cross-sections  (taken  at  right  angles  to  the  axes)  of  the 
bars,  i,  2,  .  .  .,  at  depths  dtj  d2,  .  .  .,  respectively. 

Let/i, /2,  .  .  .,  represent  the  unit  stresses  in  the  bars,  i,  2, 
.  .  .,  so  that  the  total  stresses  in  the  successive  layers  of  bars, 
i,  2,  .  .  .,  for  the  breadth,  b,  are/i  A^f^A^.  .  .,  respectively. 

Note  carefully  that  the  areas  Ait  A2)  .  .  .,  are  not  of  sections 
in  the  plane  IN  but  of  right  sections  of  the  bars.  Also  /i  is 
the  unit  stress  on  a  right  section  of  bar  i,  so  that/i  A\  is  the 
stress  in  the  bar.  Similarly  for  the  others. 

The  angles,  |3,  ft,  02,  .  .  .  (expressed  in  radians),  are  those 
made  by  the  surface,  NN',  bars  i,  bars  2,  .  .  .,  respectively, 
with  the  normal  to  the  section,  IN. 


246  APPENDIX  I 

Let  0  be  the  neutral  axis,  and  D  the  point  where  the  resultai 
C  of  the  compressive  forces  (all  acting  parallel  to  NN'),  meet 
IN. 

Let  N  0  =  kdi,  DI  =  jdi,  01  =  DI,  OL  =  »2,  etc. 

/'2V'  represents  a  section  parallel  to  //V,  Fig.  4  (7>),  and  at 
perpendicular  distance,  Ax,  from  it.     The  "  fiber,"  PPr,  paralh 
to-NN',  of  concrete,  will  be  supposed  to  have  a  section  made 
the  plane  IN,  whose  area  is  equal  to  Aa;  and  the  distance, 
will  be  called  v.     Thus  the  area  of  a  right  section  of  the  fiber  it 
Aa  cos  |8;  and,  if/ w  Jfo  wra'J  s/rew  0w  a  right  section  of  thefil 
PP',  the  total  compressive  stress  on  the  fiber  will  be  (/  Aa  cos 

After  strain,  suppose  the  plane  section,  IN,  rotates  relatively 
to  /'  N'  through  the  angle,  a,  to  /  M.    The  section,  /  M,  wi 
thus  be  supposed  to  be  plane,  as  in  the  ordinary  theory. 

For  any  fiber,  whether  of  concrete  or  steel, 

change  of  length  of  fiber 

unit  stress  =  °  °     £    .      .      (i) 

length  of  fiber 

where,    E  =  modulus  of  elasticity  of  fiber. 
Let  Es  =  the  modulus  of  elasticity  of  steel, 

Ec  =  the  modulus  of  elasticity  of  concrete, 
and, 

B* 

..-.    .    .....    0 

As  in  wedge-shaped  beams,  there  is  generally  a  stress  al 
right  angles  to  an  interior  fiber,  the  hypothesis  of  a  constant 
Ec  may  not  be  exactly  realized.  However,  any  error  from 
source  should  be  small. 

Within  working  limits   of  stress,   and  for .  the  very  smal 
values  of  a  corresponding  to  very  small  values  of  A  x,  the  change 
of  length  of  PP'  =  PQ  =  v  a  sec  0,  very  nearly;*    hence  the 


v  sin  a 

*  In  the  triangle,  P  0  Q,  by  the  law  of  sines,  P  Q  = — r.     As  a 

cos  (a  -f  ft} 

v  a. 

tends  toward  zero,  this  approaches =  v  a  sec  ft. 

cos  ft 


7] 


GENERAL   SOLUTION 


247 


units  stress,  /,  on  this  fiber,  P  P',  by  equation  i  is 

_  PQ         _  v  a  sec  0  v  a 

I ==  YJ'     c  =  Axsecp     c  =  ~Ax     c' 

Hence,  as  this  unit  stress  on  a  right  section  of  the  fiber  acts  on 
the  right  sectional  area,  A  a  cos  0,  the  total  stress  on  the  fiber  is 

—  Ec  (A  a  cos  0), 

and  its  component  perpendicular  to  IN  is 

Ec  a  cos2  0  ,        N 
(vAa). 

The  sum  of  such  components  on  the  compressed  area  of 
depth,  k  di  and  breadth,  b  is, 


,  Ecacos*0  (i  .   x 

(v  A  a)  =  -  -  ( —  b  k2  di2 ) , 

A  V.  \  o  / 


since 


A*  •*„  A 

I 

(v  A  a)  =  the  statical  moment  of   the   area  under 


compression  about,  the  neutral  axis  =  b  k  di  —  kdt. 

2 

For  the  layer  of  bars  i,  at  /,  similarly,  the  unit  stress  is 

Otherwise  in  Fig.   5,  R  Q  =  O  R  tan  a  .'.  P  Q  =  R  Q  sec  ft  =  0  R  tan  a 
sec  ft.     From  the  figure,  /3  being  constant,  as  a  approaches  zero  indefinitely, 


FIG.  5 

Q  approaches  P,  O  R  approaches  O  P  and  tan  a  approaches  or;   whence  from 
the  formula  P  Q  approaches  indefinitely, 

O  P.  a  sec  ft  =  v  a  sec  ft. 


248  APPENDIX  I 

JI  v.  a  sec  fa  Eca 

fi  =  —  Es  =  —     ——  Es  =  — -  (nvi). 
Ir  Ax  sec  fa  Ax 

Also,  as  the  area  of  a  right  section  of  bars  i,  for  the  breadth, 
b,  is  AI,  the  total  stress  in  the  bars  i,  is 

— —  (nviAi). 
Ax 

Similarly,  the  stresses  in  bars  2,  3,  .  .  .,  are 

Ec  a  Eca 

-  (nv2A2),  ~   -  (nvtAt),  .  .  . 

Ax  Ax 

As  all  the  loads  on  the  beam  were  supposed  to  act  parallel 
to  /  N}  the  part  of  the  beam  to  the  left  of  this  section  is  in  equi- 
librium under  the  loads  and  reactions  acting  on  it  and  the  internal 
stresses  along  /  N.  For  equilibrium,  the  sum  of  the  components 
of  the  stresses  perpendicular  to  IN  must  be  zero.  Therefore, 


Ax 
Eca 


COS' 


(—bk 


n  (vi  AI  cos  fa  +  v2  A2  cos  02  +  •  •  .)• 
Ax 

From  Fig.  4, 

vi  =  di  —  kdi,  vz  =  dz  —  kdi,  .  .  .; 

hence,  on  substituting  these  values,  striking  out  the  common 
factor,  and  reducing,  we  derive, 

—  cosz  $bdi2k2  +  n  di  (Ai  cos  fa  +  A2  cos  02  +  -  -  •)  k 

2 

=  n  (diAtcos  fa  +  d2A2  cos  fa  +  •  ^)    .     .      (3) 

From  this  quadratic  in  k,  the  value  of  k  is  computed,  and  thus 
k  di  =  O  N  can  be  found  and  the  neutral  axis  located.     Also, 

as  the  compressive  stresses  are  uniformly  varying,  DN  =  —  ON', 

o 

therefore,y d\  =  d\ kdi',  and 

3 


8]  RESISTING  MOMENT  249 

y  =  i--£  .      ,  .  .  •  .   (4) 

3 

8.  The  Resisting  Moment,  Ms,  of  the  Steel.  The  moment,  M, 
B,t  the  section,  IN,  due  to  the  external  forces,  is  equal  to  the 
resisting  moment  of  the  stresses  acting  along  the  section.  Call- 
ing the  perpendicular  distances  from  D  to  bars  i,  2,  .  .  .,  pi,  p%, 
.  .  .,  respectively, 

Ms=fiAipi+f2A2p2  +  .  .  .     .     .    ...     (5) 

where,  pi  =  j  di  cos  ft,  pz  =  D  L  cos  02,  etc. 

Now,  as  /i  =  —  —  nvijfz  =  —  —  nv^  .  .  .,  it  follows  that 


/»  =  -/!,/.  =  -/i,  ........  !      (6) 

fli  fli 

or  the  unit  stresses  in  the  bars  vary  directly  with  the  distances 
from  the  neutral  axis.  The  unit  stresses  in  the  interior  bars 
will  thus  always  be  less  than/i,  so  that  such  an  arrangement  of 
bars  is  uneconomical. 

On  substituting  equations  6  in  equation  5, 

ipl  +  —  A2p2  +  —A3p3+...}.    .    (7) 

Vl  Vi 

If  preferred,  after  locating  the  point  D  on  a  drawing,  the 
perpendiculars,  pi,  p2,  .  .  .,  can  be  measured  to  scale;  otherwise, 
they  may  be  computed  readily  by  the  formulas  given. 

If  the  resisting  moment  of  the  steel,  Ms,  is  less  than  that 
of  the  concrete,  MC1  for  assigned  maximum  unit  stresses,  then 
the  moment,  M,  of  the  external  forces  is  put  equal  to  the  right 
member  of  equation  7,  the  value  of  fi  ascertained,  and,  from 
equation  6,  the  values  of/2,/3,  .  .  .,  are  computed.  The  stresses 
in  bars  1,2,.  .  .,  are  thus/i  Ai,fz  A2,  .  . 

Otherwise,  if  a  certain  value  is  assigned  to  /i,  as  16,000  Ib. 
per  sq.  in.,  and  A2,  A3,  .  .  .,  are  assumed,  from  equation  7, 
AI  can  be  computed.     For  rough  computations,  Az.  A  3,  .  .  ., 
may  often  be  ignored,  in  which  case  we  can  write, 
Ms  =  /i  Ai  pi  =  fi  Aijdi  cos  ft. 


250  APPENDIX  I 

9.  The  Resisting  Moment,  Mc,  of  the  Concrete.  To  compute 
Mc,  the  position  of  the  resultant,  R,  of  the  stresses  in  the  bars, 
must  first  be  found.  The  magnitudes  of  the  forces  acting  on 
the  bars  i,  2,  3,  .  .  .,  are 


and,  as  these  are  proportional  to  /i,  the  direction  and  line  of 
action  of  R  are  the  same  for  any  value  of  /i  and  hence  for/i  =  i. 
Let  H  —  the  component  of  R  perpendicular  to  IN  when 
/i  =  i  ;  then 

?>2 

H  =  Ai  cos  ft  H  --  A2  cos  02  +  .  .  . 

9l 

Suppose  the  resultant  cuts  IN  at  G\  then,  taking  moments 
about  Z), 

H  .DG  =  Aipi  +  —  A2pz  +  —  Asp3  +  .  .  . 

Vl  Vi 

The  right  member,  presumably,  has  already  been  computed  in 
applying  equation  7;  hence  DG  is  quickly  ascertained  and 
the  point,  G,  located. 

Call  the  maximum  unit  stress  on  the  concrete  at  N,fc;  the 

unit  stress  on  the  fiber,  PP',  at  P  is  thus  —  v.    This  acts  on  the 

kdi 

area  (A  a  cos  0)  ;  hence  the  stress  on  the  fiber  is 

—  -  v  A  a  cos  j8, 
kdi 

and  the  sum  of  such  stresses  is 

/  _^kdi  f 

C  =  ^-cos$  V       (^A^)=-fj   cosp 

kdi  °  2  kdi 

Therefore,  C  =  —  fc  b  kdi  cos  0. 
2 

Taking  moments  about  G, 


Mc  =  —fcbkdi  cos2  0DG  .     ...     (8) 

2 


9,10] 


STRESSES   IN  COUNTERFORT 


251 


If,  for  assigned  maximum  values  of  /i  and  fe,  Ms  >  Me,  the 
beam  is  over-reinforced  and  M  is  placed  equal  to  the  right 
member  of  equation  8,  as  its  strength  is  now  limited  by  that 
of  the  concrete.  This  case  rarely  occurs  in  practice. 

10.  As  an  illustration  of  the  application  of  the  formulas,  take 
the  counterfort,  BM,  Fig.  6,  attached  to  the  face-slab,  NM, 
and  suppose  the  counterfort  to  be  subjected  to  a  horizontal 
earth  thrust  of  123,750  lb.,  acting  to  the  left  and  9.06  ft.  above 


Counterfort  ~          \ 

c    VV  ,  t»»\ 

^1M  11 1 A 


FIG.  6 

N,  giving  a  bending  moment  at  N  of  13,454,100  in.-lbs.  The 
section,  N  B,  corresponding  to  N I  of  Fig.  4,  is  taken  parallel  to 
the  load  (the  earth  thrust),  and  is  therefore  horizontal. 

The  width  of  the  counterfort  is  b  =  18  in.  The  inclined  bars 
have  a  total  sectional  area  =  A\  =  9.45  sq.  in.,  and  the  vertical 
bars,  a  common  area,  A  =  0.784  sq.  in.  Using  the  foregoing 
notation,  ft  =  23°  58',  ft  =  ft  =  .  .  .  =  ft  =  o.  The  dis- 
tances, d,  were  measured  from  N,  or  from  the  front  face  of  the 
vertical  slab;  di  =  128,  d2  =  108,  dz  =  100,  d*  =  92,  db  =  76, 
d*  =  60,  d7  =  44,  all  in  inches.  Assume  n  =  15. 

On   substituting  known   values   in   equation   3,   we   derive 

k  =  0.311.     Therefore,,/  =  i k  =  0.896;     whence    NO  = 

3 

kdi  =  39.8  in.,  DB  =jdi  =  114.7  m«;  also,  ND  =  —  NO  = 

3 

13  in. 

On  subtracting  N  O  =  40  from  di,  d2,  .  .  .,  we  derive  z^i, 
i>t,  etc.  The  perpendiculars,  pi,  pz,  .  .  .,  from  D  on  bars  i, 
2,  .  .  .,  are  pl  =  jdl  cos  ft  =  105,  p2  =  dz  -  ND  =  95,  .  .  ., 
respectively. 


252  APPENDIX  I 

From  equation  7  we  have,  13,454,100  =  1164/1;  whence 
/i  =  11,500  Ib.  per  sq.  in.;  from  equation  6,  fz  =  8970,.  .  ., 
/7  =  575  Ib.  per  sq.  in. 

As  the  weight  of  the  heel-slab  must  be  carried  by  the  rods, 
the  areas  and  spacing  of  the  vertical  rods  were  designed  to  carry 
their  proportionate  part  of  the  weight  of  the  heel-slab.  The 
stresses  corresponding  are  found  to  be  in  excess  of  those  due 
to  the  moment,  M.  This  excess  is  taken  up  by  the  bond  stress  in 
a  short  distance  above  N  B,  so  that,  above  a  certain  level,  only 
the  moment  stress  corresponding  to  that  level  is  carried  by  the 
vertical  rods. 

The  total  stress  in  the  inclined  rods,  fiAi  =  11,500  X  9.45 
=  108,675  Ib.,  is,  of  course,  less  than  the  stress,  127,000  Ib., 
found  by  ignoring  the  influence  of  the  vertical  rods.  This  last 
stress  is  most  easily  found  by  use  of  the  diagram,  Fig.  9,  and 
equation  u,  given  later. 

This  example  of  the  counterfort  has  been  given  more  for 
the  purpose  of  gaining  an  idea  of  the  actual  stresses  involved 
than  of  urging  the  adoption  of  the  more  refined  method  in 
practice. 

If  the  diameter  of  a  bar  is  to  be  some  multiple  of  y&  in., 
there  may  be  no  saving  by  the  use  of  the  more  exact  method. 
It  must  be  remembered,  too,  that  the  vertical  bars  are  not  always 
bonded  securely  in  the  base-slab,  the  earth  thrust  may  also  be 
much  increased  in  times  of  heavy  rains,  where  the  filling  is  not 
adequately  drained;  and,  further,  the  foundation  may  be  more 
yielding  than  estimated. 

ii.  In  the  next  example,  Fig.  7,  representing  the  heel  of  a 
T-wall  with  a  fillet,  the  wall  being  on  the  point  of  overturning, 
the  exact  method  seems  advisable. 

The  total  moment  at  the  section,  N  /,  due  to  the  two  forces 
shown,  is 

M  =  (13,000  X  2.5  +  5333  X  5)  12  =  709,992  in.-lb. 

The  reinforcement,  shown  by  the  broken  lines,  for  both 
inclined  and  horizontal  bars,  consists  of  7/8-in.  square  bars, 
8  in.  from  center  to  center,  corresponding  to  A\  =  A%  —  1.15 
sq.  in.  for  a  breadth  of  b  =  1 2  in. 


11,  12] 


HEEL  WITH  FILLET 


253 


Assuming  n  =  15,  ft  =  43°  10',  0  =  ft  =  o,  di  =  42,  dz  = 
21.5,  and  substituting  in  equation  3, 


—  b  </i2  k2  +  n  Al  d2  (cos  ft 
2 

we  find,  after  solution, 


k 


='0.238;  therefore/  =  i  --  k  —  0.921, 

o 


NO  =  kdi  =  10  in.,   DI  =jdi  =  39  in.,  ND  =  3.33  in., 
Vi=  d\—  kdi=  32,  vz=  d^—  kdi=  11.5;  therefore  —  =  0.361. 

/  ^2  \ 

By  equation  7,  Ms  =  fi  AI  [p!  -\ p2  ),pi  =  DI  cos  ft  = 

\  DI        ' 

28.4,  p2   =    l8.2; 

therefore  M  =  710,000  =  1.15  (18.2  +  6-57)/i  =  40 /r 
therefore/i  =  17,750  Ib.  per  sq.  in.,/2  =  — /i  =  6380  Ib.  per  sq.in. 

Where  the  foundation  is  good,  there  can  be  only  a  very 
small  moving  over  of  the  wall,  so  that  the  friction  force,  5333  Ib., 
at  N'  can  be  neglected  and  the  soil  reaction  included.  In  this 
particular  example  (not  given  in  full  here)  the  new  values  of /i 
and/2,  thus  found,  are  only  four- tenths  of  those  given  previously. 

12.  Special  Case.  Where  /3  =  o  and  All  the  Rods  Lie  in  One 
Plane  Which  Meets  the  Section,  IN,  in  a  Line  at  I. — Fig.  8. 
The  rectangular  section  has  the  breadth,  b  and,  as  before,  AI 
represents  the  combined  area  of  a  right  section  of  all  the  rods 


254 


APPENDIX  I 


at  I  in  the  breadth,  b.  To  agree  with  the  usual  notation,  put 
di  =  d  and  /i  =  /,  =  the  unit  stress  in  the  rods.  Equation  3 
now  reduces  to 


nd  Ai  cos  fa  k  —  nd  AI  cos  fa  =  o. 


Placing  p  = 


I  i 

=  —  ,  and  dividing  by  —  002, 


concrete  area       bd  2 

2  n  p  cos  fa  k  —  2  n  p  cos  0i  =  o, 


therefore  k=  —  n  p  cos  fa  +  V(w  p  cos  fa)2  +  2  (np  cos  fa).     (9) 

When  fa  =  o,  this  reduces  to  the  usual  formula  for  prismatic 
beams, 

(10) 


k  =  -  n  p  +  V(«  pY  +  2  (n  p) 


In  Fig.  9,  the  values  of  k  and  y  =  i k,  are  given  as 

o 

ordinates  to  the  dotted  curves  for  various  percentages  of  steel 
and  values  of  fa,  varying  from  o°  to  40°,  assuming  n  =  15. 

The  resisting  moment  of  the  steel  is  found  by  taking  moments 
about  D, 

Ms=fsAiCosfajd=fsAlpl      .     .      .    (ii) 

where  pi  =  the  perpendicular  distance  from  D  on  the  bar. 

The  resisting  moment  of  the  concrete  is  found  by  taking 
moments  about  7, 


Mc  =  —fc  b  kdj  d  =  -fc  (kj)  b 

2  2 


.        •      (12) 

For  assigned  maximum  working  values  of  fs  and  fct   the 


loop 


>]  DIAGRAM  FOR  k   AND  j  255 

ast  resisting  moment  is  equated  to  the  moment  of  the  external 

>rces. 
In  using  Fig.  9,  note  that  p  is  not  the  percentage  of  steel, 

o 

115 


90 


.20 


Percentage  Reinforcement 
FlG.  9 


256 


APPENDIX  I 


but  i/ ioo  of  the  percentage,  the  area  of  the  steel  AI,  being  the 
area  of  a  section  of  the  rods  at  right  angles  to  the  axis  and  not 
of  a  section  in  the  plane  IN. 

Resuming  the  example  of  the  counterfort,  Fig.  6,  what  will 
be  the  result  of  ignoring  the  vertical  rods?  With  AI  =  9.45  sq. 
in.  and  as  the  area  of  the  base  or  section.  N  B  =  128  X  18  = 

9-45 

2304  sq.  in.,  p  = =  0.004,  or  the  stee*  percentage  is  0.4. 

2304 

Using  Fig.  9,  with  ft.  =  24°  and  the  steel  percentage  0.4, 
j  =  0.906;  therefore./ d  =  0.906  X  128  =  116. 

Hence,  as  the  moment  of  the  earth  thrust  was  given  as  M 
13,454,100  in.-lb.,  by  equation  n, 

fs  AI  X  116  X  0.914  =  13,454,100; 

therefore  fs  AI  =  the  stress  in  the  inclined  rods  =  127,000  Ib. 
With  AI  =  9.45,  as  assumed,  fs  =  13,400  Ib.  per  sq.  in.,  in  place 


of  11,500  Ib.  per  sq.  in.,  found  before,  the  vertical  rods  being 
included. 

13.  Special  Case.  Where  the  Rods  Lie  in  One  Plane,  for 
Which  ft  =  o,  but  /3  is  not  Zero.  —  Fig.  10.  Putting  d  for  di, 
and  A  for  AI,  equation  3  reduces  to 


cos 


—  2n  A  =o. 


Therefore,  as  before,  putting  p  =  — , 

bd 


13] 


STRESSES.      SPECIAL   CASE 


k  =  —  r—  f  -  (n  p)  +  -J(np?  +  2(np)cos* 
cos2  ]8  L 


257 

.     (13) 


In  Fig.  9,  the  full  lines  give  the  values  of  k  a,ndj  =  i k, 

corresponding  to  n  =  15  and  to  various  values  of  /3  and  per- 
centages of  steel.     Equation  7,  in  this  case,  reduces  to 

M,=fsAjd (14) 

Also,  as  D  G  =  D I  =  jd,  equation  8  becomes 

'  (is) 


If,  in  Fig.   10,  we  write  p'  =  d  cos  £  =  the  perpendicular 
from  I  on  N  Nf  produced, 

#"  (16) 


The  product,  kj,  varies  with  0.  In  Table  i  the  values  of 
(kj)  cos2  j8,  are  given  for  various  values  of  p  and  /3.  The  values 
for  |8  =  o,  pertain  to  a  prismatic  beam. 

TABLE  i 
VALUES  OF  (k  j)  cos.2  ft 


100  /> 

0.2 

0.6 

I.O 

1.6 

2.0 

/?=0 

O.200 

0.303 

0-359 

0.411 

0.441 

ft  =  10° 

0.199 

0.299 

0.352 

0.403 

0.427 

/?  =20° 

0.186 

0.281 

0.329 

0.376 

0-397 

/?  =  30° 

£  =  40° 

O.l69 

0-^45 

0.251 
O.2I2 

0.294 
0.246 

0-333 
0.276 

0-351 
O.289 

It  is  seen  from  the  tabular  values  and  equation  15,  that  Mc 
as  given  by  equation  15  is  always  less  for  0  >  o  than  for  0=  o. 

In  the  ordinary  theory,  given  in  textbooks  on  "strength  of 
materials,"  for  homogeneous  "  beams  of  equal  strength"  with 
variable  depth,  vertically  loaded,  it  is  assumed  that  at  any 
vertical  section  the  theory  for  a  prismatic  beam  applies,  and 
that  the  bending  stress  at  any  point  of  the  section  acts  perpen- 
dicular to  it.  The  theory  is  thus  inadequate  to  express  the 


258  APPENDIX   I 

facts,  because  it  was  shown,  in  the  beginning  of  this  paper,  that 
the  stress  at  N,  Fig.  i,  acts  parallel  to  the  face,  NN'. 

This  common  theory,  if  extended  to  reinforced  beams  of  the 
type  shown  by  Fig.  i,  is  on  the  side  of  danger,  for  it  would  give 
for  any  0,  the  value  of  Mc  from  equation  15  corresponding  to 
0  =  o. 

The  case  where  0  =  o,  fa  =  o,  leads  to  the  ordinary  for- 
mulas for  a  prismatic  reinforced  beam,  for  which  a  valuable 
working  diagram  was  first  given*  by  Arthur  W.  French,  M. 
Am.  Soc.  C.  E. 

14.  Prismatic  Beams  with  one  layer  of  rods  parallel  to  the 
surface.    The  case  is  shown  in  Fig.  8,  when  fa  =  o.     Hence 
we  have  directly  from  (10),  (n),  (12),  the  following  formulas 
for  k,  Ms  and  Mc.    To  follow  the  usual  notation,  put  A\  = 
A  —  steel  area  in  section  IN  for  the  width  b. 


p  +  -^ 

J(np)*+2(np)     .. 

•     •    (i?) 

i                A 

3                 bd 

Mt  = 

fsAjd    . 

(18) 

s 

I 

2 

'c(kj)bd*     .     .     . 

-     •    (19) 

Writing,  Rs  =  fs  pj,  Rc  =  —fc  kj, 

Mc  =  -fckj.bd*  =  Rcbd\     .      .      .    (21) 

2 

For  n  =  15,  the  values  of  Rs  and  Rc  were  computed  for 
various  values  of  p,fs  and/,  and  plotted  as  ordinal  es  in  Fig.  n, 
the  abscissas  being  the  percentage  of  the  reinforcement  or  100  p. 
The  curves  connecting  the  points,  show  at  a  glance  the  corre- 
sponding values  of  fc  and  fs  for  a  given  p  and  R. 

Ex.  I.     Thus  for  p  =  o.oi  or  a  i%  reinforcement  and  R  —  108,  we  read 
from  the   diagram,  fc  —  600,  fs  =  12,700  Ib.  per  sq.  in.     If  fc  was  limited 

*  Trans.,  Am.  Soc.  C.  E.,  vol.  LVI,  1906,  p.  362. 


14] 


0.2  0.4 


RECTANGULAR  BEAMS 

Percentage  Reinforcement. 
0.8  1.0  1-2  1.4 


1.6  18 


259 


2.0 


0.2  0.4  0.6 


0.8  1.0  1.2  1.4 

Percentage  Reinforcement. 

JOO  |>- 

FIG.  ii 


1.6  1.8 


260  APPENDIX   I 

to  600  and  fs  to  16,000  Ib.  per  sq.  in.,  the  latter  value  for  p  =  o.oi,  would 
correspond  to  R  =  138,  fc  >  600;  .*.  the  resisting  moment  is  Mc  =  Rc  bd2 
=  1  08  bd2  and  not  Ms  =  138  bd2. 

Ex.  2.  b,  d,  p,  M,  given,  to  find  fc  and  fs.  Thus  let  b  =  8",  d  =  20", 
p  =  o.oi  i,  M  =  400,000  in.  Ibs.  .'.  R  =  M/bd2  =  125.  Find  the  inter- 
section of  a  horizontal  line  through  R  =  125  with  a  vertical  line  at  100  p  =  i.i 
and  estimate  the  ordinates,  fc  =  675,7,  =  I3»4<>o. 

Ex.  3.  Given  M  =  400,000  in.  Ibs.,/,  =  16,000,  fc  =  600.  At  the  inter- 
section of  the  curves  for  /,,  fa  the  percentage  is  0.67  (or  p  =  0.0067)  and 

M      400,000 

R  =  96.    .*.    bd2  =  —  =  —       —  =  41.70.     If  we  assume  b  =  10"    .*.    d2  = 
R  96 

417  .*.  d  =  20.4"  and  A  =  p.bd  =  0.0067  X  204  =  1.367  sq.  in.  For  this 
steel  area,  /,  and  fc  are  at  the  working  limits. 

15.  Comparison  of  maximum  unit  stresses  in  the  concrete  as 
computed  by  two  different  methods.  For  the  toe-beam,  Fig.  i, 
subjected  to  the  upward  acting  load  Q,  the  external  moment 
about  I  is  M  =  Qa  and  on  the  supposition  that  /0  is  to  be  com- 
puted as  if  ]S  =  o,  we  have  by  eq.  (15), 

2M 

=  f0  sec2  0  =  r—  —  see*  13 
d2 


where  k0,jo,  are  to  be  found  from  the  diagram  for  k,j,  Fig.  9,  for 
]S  =  o.  This  refers  to  the  state  of  stress  shown  in  Fig.  i  (i). 
For  the  state  of  stress  shown  in  Fig.  i  (2),  we  have  from  eq.  (15), 


Where  the  values  k,  j,  are  to  be  found  from  the  diagram, 
Fig.  9,  for  the  assumed  value  of  /?.  On  dividing  the  first  equa- 
tion by  the  second,  we  have, 

P*  =    kj 
fc        k0j0 
The  results  are  given  in  Table  2  for  various  values  of  8  and 

100  A. 

steel  percentages,  100  p  =  — 

bd 

Recalling  that  p0  is  always  in  excess  of  the  true  value,  the 
excess  appreciably  increasing  with  /?,  the  values  of  fc,  corre- 
ponding  to  System  (2)  of  Fig.  i,  are  seen  to  give  consistent 


15] 


COMPARISON   OF    STRESSES 


261 


results  and  in  the  right  direction.  The  regularity  of  the  changes 
in  the  ratio  is  noticeable,  and  the  near  approach  to  unity  for 
(3  =  i o  to  20  degrees. 

TABLE  2 
STEEL  PERCENTAGE,  100  p. 


100  p  = 

0.2 

0.6 

I.O 

2.0 

Po 

Po 

Po 

Po 

fc 

fc 

fc. 

fc 

/?    =    10° 

I.  O2 

I.  O2 

I.OI 

I.OO 

20° 

1.05 

1.05 

1.04 

I  .02 

30° 

I-I3 

I.  II 

1.09 

I.  06 

40° 

1.24 

1.20 

I.  17 

I.  12 

In  Fig. i,  d  =  1  N,  b  =  constant  width  of  section. 
In  Fig.  i  (i),kod  =  NO,jod  =  DI. 
InFg.  i  (2),  kd  =  N  0,jd  =  DI. 

From  the  experiments  on  dams,  previously  referred  to,  it 
was  found  that  the  vertical  unit  stress  on  horizontal  planes 
did  not  vary  uniformly,  the  stresses  near  the  down-stream  face 
being  generally  less  than  corresponds  to  the  law  of  uniform 
variation,  the  difference  being  practically  nothing  in  the  upper 
part  of  the  dam  and  increasing  as  the  foundation  is  approached, 
where  it  is  comparatively  large — considerably  over  25%. 

From  the  relation  between  this  stress  and  that  part  of  it  due 
to  flexure  alone,  we  must  infer  a  large  proportion  of  the  same 
excess  in  the  computed  over  the  real  value  of  the  stress  due  to 
direct  bending. 

A  similar  result  then  would  follow  for  the  toe-beam  just 
considered. 

In  the  case  of  the  india-rubber  and  plasticine  dams,  the  down 
stream  faces  made  angles  of  32°  and  41°  respectively  with  the 
vertical;  hence  for  values  of  /3  varying  from  30°  to  40°,  we 
should  expect /0  and  .'.  p0  for  the  toe-beam  to  be  in  excess  of  the 
true  stress  at  N,  certainly  over  25%.  From  the  ratios  in  Table 
2,  it  is  seen  that  for  such  values  of  0,  fc  is  in  excess  and  from 
previous  theoretical  considerations,  this  should  be  so  for  any 
value  of  |8  greater  than  zero.  However,  the  difference  must  be 
slight  when  /?  <  20°. 


262  APPENDIX  I 

By  reference  to  Table  2,  it  is  seen  that  p0  exceeds/,  by  from 
i%  to  5%  for  10°  <  ft  <  20°  and  by  from  6%  to  24%  for  30° 
<  |8  <4O°,  for  the  various  steel  percentages.  Since  the  junc- 
tion of  toe  and  stem  is  a  recognized  point  of  weakness  (from 
the  unequal  contraction  in  setting)  it  may  seem  desirable  to 
some  to  use  the  method  which  gives  a  decided  excess  value. 
If  this  is  to  be  done,  then  the  procedure  is  very  simple.  Thus 
if  po  is  assigned  a  limit  of  say  650  lb./in.2,  then/0  =  650  cos2  0; 
with  this  value  and  the  maximum  unit  stress  allowed  in  the 
steel,  the  diagram  for  prismatic  beams,  Fig.  n,  can  be  con- 
sulted and  the  solution  effected  as  usual.  As  a  matter  of  fact, 
the  maximum  unit  stress  in  the  concrete  of  toe-beams,  as  usually 
designed,  is  very  small,  the  steel  being  the  determining  factor, 
so  that  the  final  results  will  not  differ  very  appreciably,  which- 
ever method  is  adopted. 

1 6.  Shear  in  Wedge-Shaped  Reinforced  Concrete  Beams. 
Consider  the  toe,  Fig.  12,  subjected  to  the  vertical  load  Q,  the 
stress  C,  inclined  at  the  angle  0,  the  tension  T  in  the  rod  at  / 
inclined  at  the  angle  /?o  to  the  horizontal,  and  the  shear,  V, 
in  the  section  N  /,  to  be  .carried  by  the  concrete,  in  addition  to 
the  vertical  component  of  C  already  included.  V  is  thus  the 
shear  in  the  section  still  to  be  accounted  for  after  the  vertical 
components  of  C  and  T  have  both  been  considered.  For  equi- 
librium, we  must  have, 

V  =  Q  -  C  sin  j8  -  T  sin  00  .      .      .      -    (22) 

This  portion,  LMNI,  of  the  beam  acts  on  the  part  of  the  beam 
to  the  right  of  the  section,  NIt  with  forces  exactly  equal  and 
opposed  to  C,  T,  and  V. 

When  the  steel  is  horizontal,  j8o  =  o  and  V  =  Q  —  C  sin  /3. 
This  refers  to  the  case  shown  in  Fig.  1(2).  For  the  case  shown 
in  Fig.  i  (i),  C  is  horizontal  and  V  =  Q.  When  T,  Fig.  12, 
acts  upward,  the  last  term  in  the  first  formula  changes  sign. 
To  make  an  exact  analysis,  consider  Fig.  13,  representing  a 
portion  of  the  beam  between  the  two  vertical  sections,  IN  and 
I'N',  dx  apart  and  subjected  to  the  forces,  F,  F',  C  and  C' 
(the  horizontal  components  of  which  are  Co,  Co),  T,  the  re- 


16] 


SHEARING   STRESS 


263 


sultant  of  fhc  tensions  in  the  bars,  //',  LL',  SSf,  acting  at  G 
and  inclined  at  the  angle,  ft0,  to  the  horizontal  and  the  corre- 
sponding resultant,  Tf,  at  £',  supposed  to  act  in  the  line  of  T, 


FIGS.  12  AND  13 

but  opposite  in  direction.  The  constant  .width  of  the  cross- 
sections  will  be  called  b,  and  the  unit  shear  on  a  horizontal  plane, 
A  A',  just  below  the  neutral  surface,  v.  We  have, 

I'D'  -  ID  =  j  (I'N'  -  IN)  =  j  dx  (tan  ft  +  tan  ft) 
DE  =  IE  -  ID  =  I'D'  -  dx  tan  ft  -  ID  =  j  dx  (tan  ft 

+  tan  fti)  —  dx  tan  ft  .     .     /     .      .    (23) 

Now,  whatever  the  inclinations  assumed  of  C  and  C",  we  have 

C0  =  T  cos  fto (24) 

CY  —  Co  =  v  b  dx (25) 

the  latter  equation  resulting  from  placing  the  sum  of  the  horizon- 
tal components  of  the  forces  acting  on  A  A'  N'N  equal  to  zero. 
Assume  the  state  of  stress  shown  in  Fig.  i  (2),  so  that  C  and  C' 
are  inclined  to  the  normal  to  IN  at  the  angle  0;  whence,  Co 
=  C  cos  ft,  C'o  =  C'  cos  ft,  andy  is  to  be  found  by  aid  of  (3)  or 
the  diagram  Fig.  9,  for  the  given  value  of  ft. 


264  APPENDIX   I 

On  taking  moments  of  the  forces  acting  on  ±l'NrN  about 

C,  we  have, 

Co'  (GD  +  DE)  -CQ.GD-V'.dx-C'  sin  0  dx  =  o 
.'.  (Co'  -  Co)  GD  =  (V  +  C  sin  0)  dx  -  C0' .  DE 

substituting  the  values  of  (Co'  —  Co)  and  DE  from  (25)  and 

(23)  and  dividing  by  dx, 

vb.GD=  (V  +  C'  sin  0)  -  CY  [/  (tow  0  +  ton  0^  -  /aw  AJ 
On  taking  the  limit  as  dx  approaches  zero,  so  that  C0'  is  re- 
replaced  by  C0,  C'  by  C  and  V  by  V  =  Q  -  T  sin  00  -  C  sm  0, 

from  (22),  we  obtain, 

fl&  .  GD  =  Q  +  (Co  tan  fa  -  T  sin  00)  -  j  C0  (tow  0+tow  A)  (26) 
If  we  denote  by  M  the  moment  of  the  external  forces  about 

any  point  of  the  section  IN,  we  have,  taking  moments  in  turn 

about  G  and  D, 

M  =  Co . GD  =  T  cos  0o  .CD; 
M  M 

•     r*     __  'T»  

~  CD'        CD  o?s  0o* 

On  substituting  these  values  in  (26),  it  can  be  written  in 
the  form,  more  convenient  for  computation, 

Mr  -i 

vb.GD  =  Q j  (tan  0  +  tan  fa)  -  tan  fa  +  tan  00    .    (27) 

CDi— 


in  which  Q  is  the  algebraic  sum  of  the  loads  and  reactions  (if 
any)  to  the  left  of  the  section. 

For  one  layer  of  bars  at  II',  00  =  fa,  GD  =  ID  =j  .  IN, 
Co  tan  0i  =  T  sin  00,  by  (24),  (26)  and  (27)  reduce  to, 

vb  .  GD  =  Q  -  j  (C  sin  0  +  T  sin  fa)       .      .    (28) 
vb.j  .IN  =  Q-—-(tan  (3  +  tan  fa)  .      .    (29) 

The  latter  is  the  most  convenient  form  for  computing  v.  The 
value  of  v  is  found  from  any  one  of  the  equations  (2 6) -(2 9)  on 
dividing  by  b  .  GD. 

From  (28),  it  is  observed,  that  in  the  right  member,  in  place 
of  subtracting  the  sum  of  the  vertical  components  of  C  and  T 
from  Q,  that  onlyj  times  this  amount  is  subtracted. 


16,17]  BOND   STRESS  265 

When  the  state  of  stress  represented  in  Fig.  i  (i)  is  assumed, 
C  and  C"  being  normal  to  IN  and  j,  termed  now  jQ,  being  com- 
puted as  if  j8  was  zero,  the  computation  proceeds  as  before  and 
it  will  be  found  that  the  only  change  in  formulas  (27)  and  (29) 
is  that^'o  replaces  j,  noting  too,  that  GD  in  (27)  is  dependent 
on  j0. 

For  the  common  case  of  one  layer  of  bars,  the  right  member 
remains  the  same  for  both  cases,  the  left  members  are,  vbj .  IN 
and  v0  bjo .  IN,  for  the  two  cases,  where  v0  is  the  unit  shear 

1)  JQ 

for  the  system  of  stress  Fig.  i  (i) ;  whence,  —  =  — .    The  latter 

z>o      J 

ratio  being  very  near  unity  (see  Fig.  9),  the  values  v  and  VQ  are 
practically  the  same.  The  variation  of  shear  over  the  cross- 
section  will  be  discussed  later. 

17.  Bond  Stress.  Let  «i,  u2,  u3,  .  .  .,  denote  the  unit  bond 
stresses  in  the  bars,  //',  LL',  SS',  .  .  .,  and  20i,  202,  203,  .  .  ., 
the  areas  per  linear  inch  of  surface  of  the  bars,  for  the  width  b 
for  the  respective  bars;  then  since  the  shear  on  the  horizontal 
plane  A  A',  equals  the  sum  of  the  horizontal  components  of  the 
bond  stresses  in  the  bars, 

b  v  dx  =  HI  IT  cos  ft  20i  +  uz  LL'  cos  02  202  +  .  .  . 
=  dx  (u^i'-^Uz  202  +  .  .  .)• 

Now,  the  unit  bond  stress  in  any  rod  is  proportional  to  the 
unit  elongation  of  the  rod,  or  to  its  unit  stress,  which  varies 
with  the  distance  from  0  (see  eq.  6) ;  hence, 

OL  OS 

U*  =   —  «1,    "3    ==    —  «1, (30) 

Therefore  the  previous  equation  can  be  written,  on  dividing 
by  dx, 

OL  OS  -. 

i  +  — y  202  +  —  203  +  .    .    .   \  =  Vb  .     .      (31) 

On  writing  for  vb,  its  value  given  by  (27), 

r  OL  OS  -i 

ui  [_20i  +  — -  202  +  —  203  +  -  -  J 


266  APPENDIX  I 


=  £~  \Q  -  —  [j  (ton  ft  +  ton  181)  -  tan  &  +  tan  00]  J     (32) 


From  this  equation  «i  can  be  found  and  then,  by  aid  of  (30), 
MZ,  u3)  .  .  .,  can  be  computed. 

For  only  one  layer  of  bars  at  II',  vb  is  given  by  (29)  and, 

Ui  =  vb  -f-  Z0i,  or, 

Ul=j.  IN  aj  Q  ~  7F(fam  *  +  *"*  ft)  t  '  '  (33) 

If  the  state  of  stress,  Fig.  i  (i),  is  assumed,  then  J  is  to  be 
replaced  by  j0  in  the  last  two  equations  with  a  corresponding 
change  in  G  D  in  (32). 

In  all  the  formulas  (26X33),  ^  ^"  *s  directed  to  the  right 
and  upward,  sm  /30,  tow  /30,  are  to  be  replaced  by  (—sin  /30), 
(—  tan  j80),  observing  too,  that  for  one  layer  of  bars,  00  =  fti- 

For  the  case  of  many  layers  of  bars,  the  value  of  G  D  can  be 
found  as  in  App.,  Art.  9  and  the  direction  of  T  or  value  of  00, 
by  drawing  the  polygon  of  forces  proportional  to  the  tensions 
in  the  bars: 

A        °L    A        °S    A 

"  01     *'  01     "  ' 

1  8.  Ex.  I.  As  an  application,  let  it  be  proposed  to  find  the  unit  shear 
and  bond  stresses  for  the  heel-slab  with  the  fillet,  App.,  Fig.  7,  Art.  9  al- 
ready considered. 

We  have  given,  Q  =  18,333  Ibs.,  M  =  710,000  in.  Ibs.  The  tension  in 
the  bar  which  makes  the  angle  /3i  =  43°  10'  with  the  horizontal  =  f\A\  and 

that  in  the  horizontal  bar  =  /i  —  —  A2  =  fi  —  —  A2  =  0.361  A2.    Hence,  since 

AI  =  A2  in  this  instance,  these  tensions  are  in  the  ratio  i  :  0.361;  therefore 
at  the  point  of  intersection  of  the  two  bars  on  the  figure,  Conceive  a  force  i 
acting  along  bar  i  and  a  force  0.361,  acting  along  bar  2  and  by  construction 
find  their  resultant,  whose  direction  is  that  of  T.  This  resultant  is  found 
to  make  the  angle  /90  =  32°  05'  with  the  horizontal  and  to  cut  N  I  at  a  point 
G  such  that  G  D  =  32.1  inches.  These  same  results  can  be  found  analytically. 
On  substituting  in  (27),  b  =  12,  G  D  =  32.1,  j  =  0.921  (Art.  9),  M  = 
710,000,  tan  0i  =  0.938,  tan  /80  =  0.627,  tan  /3  =  o,  we  have, 

12  X  32-i  »  =  i8,333  --        -  (0.921  X  0.938  —  0.311) 


18] 


EXAMPLES 


267 


whence,  v  =  15.9  lb./in.2,  is  the  horizontal  unit  shear  between  the  neutral 
axis  and  the  horizontal  bar. 

To  compute  bond  stress,  note  that  both  the  inclined  and  horizontal  re- 
inforcement consists  of  %  in.  square  rods,  spaced  8  in. — center  to  center. 
Therefore  for  a  breadth  b  =  12  in., 


I,  01    =    S  02 

We  have  by  (31),  using  the  value  of  v  just  computed, 


-  (3-5)  =  5-25. 

o 


.25  «!   [  I  +  — j-J  =  v  b  =  15.9  X  12 


.'.  «i  =  26.7  lb./in.2,  «2  =  —  —  «i  =  9-6  Ib.in.2, 

the  unit  bond  stresses  in  bars  I  and  2. 

Ex.  2.     For  an  application  of  (29)  and  (33)  to  the  toe  of  a  retaining  wall, 
consider  the  toe  with  a  fillet  Fig.  14,  subjected  to  an  upward  soil  reaction  of 


10  100  Ibs. 


FIG.  14 


10,100  Ibs.  for  b  =  12",  acting  1.59'  to  the  left  of  the  vertical  section  7^. 
For  a  steel  percentage  about  o.i,  from  Fig.  9,  for  /3  =  45°,  j  —  0.92;  hence 
assuming  /,  =  16,000  lb./in.2,  by  (14),  we  find  the  corresponding  area  of 
metal  for  b  =  12". 

M  10,100  X  1.59  X  12         192,900 

=  0.29  sq.  in. 


A  = 


fs.j  d       16,000  X  0.92  X  45-5    '  670,000 
As  this  is  very  small,  take  %"    D    bars,  i2"  c  to  c,  the  bars  being  placed 
2.5  in.  from  bottom  of  toe,  giving  A  =  0.39,  S0i  =  2.5. 
We  have  here,  ft  =  o,  ft  =  45°, 

192,900 


IO,IOO  — 


45.5 


=  5870 


Formulas  (29  and  (33),  can  be  written,  putting  IN  =  d, 
Ui  Jd  Z0i  =  Q  —  —  tan  /3  =  vb  .jd 


5870 


0.92  X  45.5  X  2.5        105 


268  APPENDIX  I 

The  unit  shear  v  is  easily  found  from  the  unit  bond  stress  u\. 

Ui  S0i       56  X  2.5  , 

v  =  —  —  =  -  -  =  12  lb./in.2 

b  12 

The  values  of  u\  and  z>  are  well  within  safe  limits. 

In  the  solution  above,  for  safety,  the  friction  on  the  base  and  the  weight 
of  toe  were  neglected,  because  the  junction  of  the  upper  part  of  the  toe  with 
the  vertical  slab  is  a  recognized  point  of  weakness,  from  the  unequal  contrac- 
tion of  the  concrete  in  drying  out.  Besides  the  outer  surfaces  of  both  toe  and 
stem  are  in  compression.  The  latter  objection  does  not  hold  where  the  heel 
and  stem  unite.  For  abundant  security,  some  designers  ignore  the  fillet  of 
the  toe  altogether  in  the  computation  of  Ui,  v  and  A\t  but  this  plan  seems 
unnecessarily  wasteful  in  steel. 

For  the  commonly  recurring  case  of  the  toe,  Fig.  i,  where 
ft  =  o,  (29)  and  (33)  can  be  written  in  the  compact  form, 

u\  .  jd  Z0i  =  Q  --  tan  0  =  vb  .  jd  .      .      .    (34) 
d 

in  which  Q  and  M  are  the  shear  and  moment  at  the  section  IN, 
due  to  the  external  forces. 

For  a  prismatic  beam  0  =  o 

.'.  ui  .jd  20!  =  Q  =  vb  .jd    ....    (35) 

19.  Variation  in  Shear  Over  Section.  The  unit  shear  q  on 
a  horizontal  (or  vertical)  plane  at  the  distance  yi  above  O  in 
the  case  of  the  toe,  Fig.  i  (i),  for  the  state  of  stress  assumed 
in  that  figure,  can  be  found  by  a  method  similar  to  that  used 
for  prismatic  beams,  only  the  rise  of  N'  above  N  and  of  0'  above 
0,  Fig.  13,  must  be  considered  in  the  limits  of  the  integrations. 
As  the  work  is  long,  only  the  final  result  will  be  given.  As 
hitherto,  b  =  constant  breadth,  d'=  depth  NI,  M  =  moment 
of  external  forces  about  /  and  &o,  Jo,  refer  to  the  state  of  stress 
shown  in  Fig.  i  (i),  where  Co  is.  computed  as  if  the  beam  were 
prismatic  and  of  the  constant  depth  d. 

The  value  of  q  is  given  by  the  following  equation,  in  which 
Q  =  external  shear  at  section  IN: 


.(36) 


k<?d  (Qd  -  M  tan  (3)  +  [2  (i  -  k0)  M  tan  ft] 
M  tan  ft 


19]  VARIATION   IN   SHEAR   OVER   SECTION  269 

The  unit  shear  at  0  is  found  by  putting  y\  =  o, 

M 

Q  --  tan  ft. 

q  (at  O) 


o  .j 

This  is  the  value  of  v,  found  from  (29),  on  changing  j  to  jo  to 
pertain  to  the  state  of  stress  assumed. 

At  N,  the  shear  is  found  by  putting  yi  =  kad. 

2  M  tan  13 

q  (at  N)  =  -r-r-nr  =  /o  tan  ft, 
k0jo  od* 

as  can  be  proved  independently  by  considering  the  conditions 
of  equilibrium  of  an  infinitesimal  wedge  at  N. 

Another  check  is  afforded  by  taking  ft  =  o,  whence 


which  is  easily  proved  independently. 

In  Fig.  i  (i),  Q  may  represent  the  soil  reaction  (friction 
omitted),  which  acts  at  the  distance  a  to  the  left  of  IN\  whence 
M  =  Qa. 

On  dividing  eq.  (34)  by  (k0y0  bd3),  it  can  be  put  in  the  form, 

q  =  I  +  m  yi  +  n  y?, 

where  q  and  y\  are  the  variables,  /  and  m  are  both  positive, 
though  n  may  be  positive  or  negative. 

When  n  =  o  this  is  the  equation  of  a  straight  line,  q  increas- 
ing with  yt.  When  n  is  not  zero  it  is  the  equation  of  a  parabola. 
Let  d'  —  vertical  depth  down  to  the  steel,  of  the  left  or  truncated 
end  of  the  toe  and  suppose  Q  to  act  approximately  at  the  center 
of  the  base,  corresponding  to  a  uniformly  distributed  soil  reac- 
tion; then  it  can  be  shown  that  n  >  o,  =  o,  <  o,  according 

d      d  d  d 

as  d1  <  -,  =  -  or  >  -  and  that  q  increases  with  yi  for  d'  <  - 

333  3 

or  dr  =  -,  but  for  d'  >  —  q  may  first  increase  and  then  decrease 
3  3 


270  APPENDIX  I 

or  it  may  increase  throughout  or  decrease  throughout,  as  yi 
varies  from  o  to  ON.  For  all  cases,  the  value  of  q  at  N  is 
/o  tan  j(3. 

It  may  be  well  to  remark  that  the  unit  shear  on  a  plane 
inclined  to  the  horizontal  may  be  greater  than  the  shear  on  a 
horizontal  plane,  since  there  is  generally  a  vertical  component 
of  stress  to  be  considered  in  wedge-shaped  beams. 

The  general  formula  above  refers  to  the  state  of  stress  shown 
in  Fig.  i  (i).  If  the  state  of  stress  of  Fig.  i  (2)  is  assumed 
the  same  formula  (34)  is  derived,  except  that  ko,j0,  are  to  be 
replaced  by  k,  j,  which  agrees  with  previous  results  for  yi  =  o. 

Reasons  have  been  given  in  App.,  Art.  3,  for  the  belief  that 
the  linear  law  of  variation  of  stress  shown  in  Fig.  i,  (i)  or  (2), 
is  only  approximately  true.  It  follows  that  the  variation  in 
shear,  as  given  by  (36),  which  is  based  on  this  law,  is  only  approxi- 
mately true,  the  actual  shear  at  the  surface  being  less  than  the 
equation  indicates. 

20.  Spacing  of  Bars.  It  is  a  great  convenience  to  have  at 
hand  tables,  giving  the  spacing  of  round  and  square  bars  for 
various  areas  of  metal  for  12"  width,  such  as  is  given  in  Marsh 
and  Dunn's  "  Reinforced  Concrete  Manual,"  pp.  238-241;  but 
lacking  such,  a  little  computation  is  necessary. 

Thus,  for  one  layer  of  bars,  let  A  —  cross-section  of  metal 
in  width  b,  x  =  spacing  of  bars  or  distance  from  center  to  center 
of  bars,  each  one  having  the  cross-sectional  area  a.  Then  con- 
ceiving the  metal  in  a  continuous  sheet, 

A         a 

—  =  —  =  area  in  i  unit  width. 
b          x 

Given  three  of  the  quantities  A,  a,  b,  #,  the  other  can  be 
found  from  this  equation. 

Thus  if  A  =  1.69  sq.  in.,  for  b  =  12"  and  we  wish  the  spac- 

ing of  y*J'  square  bars  .'.  a  =  —  sq.  in. 

16 


* 

a  b       16 

x  =  —  =  -          —  =4  inches  spacing. 
A         1.69 


20,  21] 


RADIUS   OF   BENT  BARS 


271 


20  0 

Similarly  in  bond  stress,  using  previous  notations,  —  =  —  = 

area  of  surface  of  bars  per  linear  inch  for  unit  width. 

Thus  the  spacing,  x  inches  for  a  bond  stress,  Ui  =  80  Ibs. 

sq.  in.,  is,  x  =  — ;  or  since  by  Art.  18,  Ex.  2  for  one  layer  of  bars, 
20 

I  Ui  OU\  0 

—  =  —  .'.  x  =  —  =  80  — ,  where  v  =  unit  shear  computed 
20        vb  v  v 

from  (29). 

i        Uijd  Sob.jd.o. 

For  prismatic  beams,  by  (35),  —  =  — —  .".  x  =  - 

2o         Q  Q 

21.  Compressive  Stress  in  Concrete  Due  to  Bending  a  Re- 
inforcing Rod  Under  a  Tension  T.    Suppose  a  square  rod  of 


FIG.  15 


diameter  b  to  be  bent  to  a  circular  arc  of  radius  r.  Consider  a 
small  portion  of  this  arc,  with  a  central  angle  dB  and  length, 
rdd  and  let  the  unit  stress  in  the  bent  rod  be/,  whence  the  total 
stress  in  it  =  T=fb2.  See  Fig.  15.  The  bent  rod  presses 
the  concrete,  the  area  of  contact  being  b  r  d  6  and  if  p  =  unit 
compression,  the  total  resisting  force  of  the  concrete  is,  P  = 
pbrdd  (de  being  very  small).  The  forces  T,  T,  P  of  the  figure 
are  in  equilibrium,  hence  balancing  components  parallel  to  P, 

de 

pbrdd  =  2  T  sin  —  =  T  dd  =  fb2 .  dd 

2 


272  APPENDIX  I 

Here,  the  sine  has  been  replaced  by  the  arc,  since  do  can  be 
made  as  near  zero  as  one  pleases. 
Hence  the  formula, 

Pr=fb       ......    (37) 

If  we  take/  =  16,000  Ibs.  sq.  in.  and  r  =  25  b, 

16,000 

p  =  -       -  =  640  Ibs.  sq.  in., 
25 

which  is  about  the  working  limit  of  concrete  in  compression. 
The  radius  should  not  be  less  than  25  diameters  of  the  square 
rod  and  preferably  more.  Sharp  bends  should  always  be  avoided 
or  rupture  may  ensue. 

If  the  rod  is  round,  of  diameter  b,  let  the  effective  area  be 
taken  roughly,  the  same  as  before. 

b2 

:.  pbrde  =  T  d6  =  f  TT  —  de 

4 

pr=o.7S$fb   .......    (38) 

If  r  =  2ob,f=  16,000,  p  =  628;  so  that  about  20  diameters 
of  the  rod  is  the  minimum  radius  that  should  be  used  in  the 
case  of  bent  round  rods.  Similarly,  when  a  straight  rod  under  a 
stress  of  16,000  Ibs.  sq.  in.,  is  fastened  by  screw  and  nut  to  a 
flat  plate,  bearing  against  concrete,  the  bearing  area  of  the  plate 

should  be  at  least,  -       -  =  25  times  that  of  the  steel. 
650 

22.  Length  of  Bar  to  Embed  in  Concrete.  Let  us  use  a 
working  bond  stress  for  plain  bars  of  80  Ibs.  sq.  in.,  which  is 
about  one-third  of  the  ultimate. 

With  a  square  bar  of  diameter  d,  let  I  =  length  to  embed 
to  develop  the  required  bond  strength,  corresponding  to  a 
tensile  unit  stress  of  16,000  Ibs.  sq.  in.  The  area  of  the  surface 


4  dl  X  80  =  16,000  dz  .'.  /  =  50  d. 


22,23]  WORKING   STRESSES  273 

For  round  bars  of  diameter  d,  we  have,  similarly,    TT  d  I  X  80  = 
d2 

l6,OOO  7T  .".    /   =    5O  d. 

4 

Hence  for  plain  round  or  square  bars,  an  embedment  of  50 
diameters  is  safe.  For  deformed  bars,  allowing  a  bond  stress 
of  100  Ibs.  sq.  in.,  an  embedment  of  40  diameters  is  allowed. 
This  embedment  will  likewise  suffice  for  plain  round  or  square 
bars  with  hooked  ends,  bent  180°  to  a  radius  of  3  diameters 
with  a  short  length  of  bar  beyond  the  bend. 

23.  Working  Stresses.  Although  the  elastic  limits  of  soft, 
mild  and  hard  steel  may  average  33,000,  37,000  and  55,000 
Ibs.  sq.  in.,  respectively,  yet  the  modulus  of  elasticity  of  either 
kind  of  steel  is  about  30,000,000  Ibs.  sq.  in.,  which  is  10  to  15 
times  that  for  concrete.  It  follows,  if  the  steel  in  a  beam  is 
stressed  to  3000  Ibs.  sq.  in.,  since  the  surrounding  concrete 
stretches  as  much  as  the  steel,  the  stress  in  the  concrete  is 

3000 

=  200  Ibs.  sq.  in.,  when  the  modulus  for  the  steel  is  15 

'5 
times  that  for  the  concrete. 

As  this  is  about  the  limit  for  the  tensile  stress  of  the  concrete, 
as  the  stress  in  the  steel  increases,  invisible  cracks  occur  which 
become  visible  at  10,000  to  16,000  Ibs.  sq.  in.  stress  in  the  steel,* 
so  that  it  is  regarded  undesirable,  for  any  kind  of  steel,  to  exceed 
the  limit  16,000.  We  thus  gain  no  advantage,  within  wqrking 
limits,  in  using  the  hard  steel,  in  spite  of  tlie  fact,  that  its  elastic 
limit  and  breaking  strength  are  so  much  greater  than  for  mild 
steel. 

As  giving  a  resume  of  the  practice  of  the  present  day,  the 
reader  is  referred  to  the  "  Report  of  Special  Committee  on 
Concrete  and  Reinforced  Concrete,"  published  in  Proceedings 
Am.  Soc.  C.  E.  for  Feb.,  1913  (or  Trans.,  1914). 

A  few  of  the  recommendations  will  be  given  here  that  apply 
more  particularly  to  the  design  of  the  slabs  of  reinforced  concrete 
retaining  walls. 

*  See  Turneaure  &  Maurer's  "Reinforced  Concrete,"  p.  40. 


274 


APPENDIX  I 


The  committee  advises  that  the  following  values  of  the 
ultimate  compressive  strength  of  concrete,  in  pounds  per  square 
inch,  should  be  the  maximum  values  allowed  for  static  loads: 


Aggregate 

1:1:2 

i:i#:3 

1:2:4 

1:2^:5 

1:3:6 

Granite,  trap  rock 

•I    -J.QQ 

2  8OO 

2  2OO 

I  8OO 

T   AOO 

Gravel,  hard  limestone  and   hard 
sandstone  ...    . 

3  ooo 

2  5OO 

2  OOO 

I  6OO 

T   ^OO 

Soft  limestone  and  sandstone  .... 
Cinders  

2,200 

800 

1.800 
7OO 

1,500 
6OO 

1,200 

COQ 

I.OOO 
4OO 

"  The  extreme  fiber  stress  of  a  beam,  calculated  on  the  as- 
sumption of  a  constant  modulus  of  elasticity  under  working 
stresses,  may  be  allowed  to  reach  32.5%  of  the  compressive 
strength."  For  a  2000  Ibs.  per  sq.  in.  ultimate  strength,  this 
would  limit  fc  to  650  Ibs.  sq.  in. 

In  beams,  where  the  computed  maximum  shearing  stress 
is  used  as  the  means  of  measuring  the  diagonal  tension  stress, 
the  maximum  shearing  stress  may  be  2%  of  the  ultimate.  This 
refers  to  prismatic  beams  without  web  reinforcement.*  For  the 
2000  ultimate,  the  limiting  value  of  v  is  2%  of  2000  =  40  Ibs. 
sq.  in. 

For  concrete  in  shear  not  combined  with  tension  or  com- 
pression-punching shear — 6%  of  the  ultimate  or  120  Ibs.  sq.  in. 
for  the  2000  ultimate  may  be  used. 

The  unit  bond  stress  between  concrete  and  plain  reinforcing 
bars,  is  allowed  to  reach  4%  of  the  ultimate  or  80  Ibs.  sq.  in. 
for  an  ultimate  of  2000  Ibs.  sq.  in. 

The  committee  does  not  give  figures  as  to  deformed  bars, 

*  The  diagonal  tension  stress  is  a  function  of  the  shear  and  moment,  though 
experiments  on  prismatic  beams  seem  to  indicate  that  shear  is  the  chief  factor, 
which  fact  is  provisionally  assumed  in  dealing  with  the  toes  and  heels  of  the 
walls  of  Chap.  IV,  for  which  experimental  data  are  lacking. 

For  wedge-shaped  beams,  the  unit  shear  at  the  neutral  axis  is  smaller  than 
for  the  enclosing  prismatic  beam,  and  it  is  possible  that  shear  is  not  to  the  same 
extent  "to  be  used  as  the  means  of  measuring  diagonal  tension."  Perhaps  a 
value  of  the  unit  shear  intermediate  between  that  for  the  toe  and  prismatic 
beam  may  be  more  suitable  for  use  with  the  assigned  limit  to  the  unit  shear,  as 
to  which,  at  present,  the  best  judgment  of  the  engineer  must  be  exercised. 


23]  WORKING  STRESSES  275 

which  are  said  to  have  a  bond  stress  (for  the  2000  ultimate)  of 
80-150  Ibs.  sq.  in.;  neither  do  they  refer  to  the  usual  increase 
in  bond  stress  to  100  Ibs.  sq.  in.  for  plain  bars  hooked  properly 
at  the  ends. 

The  modulus  of  elasticity  of  concrete  has  a  wide  range.  If 
n  =  ratio  of  modulus  of  steel  to  that  of  concrete  and  s  =  ulti- 
mate compressive  stress  in  concrete,  then  it  is  recommended 
in  computations  for  locating  the  neutral  axis  or  finding  the 
resisting  moment  in  beams,  to  take, 

n  =  15,  when  s  is  2200  Ibs.  sq.  in.  or  less, 
n  =  12,  when  s  >  2200  but  <  2900  Ibs.  sq.  in., 
n  =  10,  when  s  is  greater  than  2900  Ibs.  sq.  in. 
The  lateral  spacing  of  parallel  bars  should  not  be  less  than 
three  diameters,  from  center  to  center,  nor  should  the  distance 
from  the  side  of  the  beam  to  the  center  of  the  nearest  bar  be 
less  than  two  diameters. 

Reinforcement  for  shrinkage  and  temperature  stresses  should 
not  generally  be  less  than  1/3  of  i%. 

In  the  case  of  continuous  beams  and  slabs,  the  bending 
moment  at  the  center  and  at  support  for  interior  spans  shall  be 
taken  at  (1/12)  wl2,  where  w  represents  the  load  per  linear  foot 
and  I  the  length  in  feet,  of  the  span. 


APPENDIX  II 
Experiments  on  Retaining  Walls 

The  following  data,  referring  to  model  retaining  walls  at  the 
limit  of  stability,  are  taken  from  a  paper  by  the  author  on 
"  Experiments  on  Retaining  Walls  and  Pressures  on  Tunnels,"* 
to  which  the  reader  is  referred  for  many  additional  experiments 
and  a  much  fuller  discussion. 

As  hitherto,  w  and  wf  represent  the  weight  in  pounds  per 
cubic  foot  of  filling  and  wall  respectively,  <p  =  angle  of  repose 
of  filling,  $  =  angle  of  friction  of  filling  on  wall.  For  the 
vertical  rectangular  walls,  Figs.  16,  17,  18,  19,  h  =  height,  t  = 
thickness  of  wall  in  feet.  The  earth  thrust  against  the  wall 
is  supposed  to  act  at  one-third  the  height  from  the  base  of  the 
wall  to  the  surface  of  the  filling. 

In  the  first  four  designs,  where  the  surface  of  the  filling  was 
horizontal,  the  centers  of  pressure  on  the  base  or  base  produced, 
were  found,  (i)  by  the  Rankine  theory,  in  which  the  earth 
thrust  is  supposed  to  act  horizontally  and  (2)  where  this  thrust 
is  supposed  to  be  inclined  at  the  angle  <pf  below  the  normal  to 
the  wall.  On  combining  these  thrusts  with  the  weight  of  wall, 
as  usual,  the  resultant  is  found  to  strike  the  base  or  the  base 
produced,  at  R  in  the  first  case  (Rankine's  theory)  and  at  / 
by  the  second  method.  The  earth  thrust  in  case  (i)  was  com- 
puted by  Rankine's  formula  (Art.  48) ,  in  case  (2),  by  the  formulas 
of  Art.  44,  which  include  the  influence  of  the  wall  friction. 

For  brevity  let, 

q  =  distance  from  center  of  pressure  on  the  plane  of  the  base 
to  the  outer  toe,  divided  by  t,  using  the  theory  that 
rationally  includes  the  whole  of  the  wall  friction; 


*  Trans.  Am.  Soc.  C.  £.,  vol.  LXXII,  p.  403  (1911). 
276 


EXPERIMENTS 


277 


qQ  =  same,  according  to  the  Rankine  theory,  q  and  q0  will  be 
taken  as  positive  when  the  resultant  on  the  base  cuts  it 
within  the  base;  otherwise  negative. 

Fig.  16  represents  Lieut.  Hope's  wall  of  bricks  laid  in  wet  sand, 
backed  by  sand  level  with  its  top:  h  =  10,  t  =  1.92,  w  =  95.5, 
w'  =  100,  <p  =  (p  —  36°  53'.  The  wall  was  20  ft.  long.  Ignor- 


><  0.159'>> 


FIG.  1 8 


FIG.  19 


ing  any  leaning,  q  =  +  0.04,  go  =  —  0.58;  but  if  we  reckon  the 
probable  overhang  at  the  moment  of  failure  as  4  in.,  it  is  found 
that  q  =  —  0.02. 

Fig.  17  shows  Baker's  wall  of  pitch-pine  blocks,  backed  by 
macadam  screenings,  the  level  surface  of  which  was  0.25  ft. 
below  the  top  of  the  wall:  h  =  4,  /  =  i,  w  =  101,  wr  —  46, 
<p  =  39°,  <p'  (assumed)  =  22°,  q  =  —  0.06,  q0  =  —  0.55. 

Trautwine's  experimental  wooden  wall  is  shown  in  Fig.  18. 
Only  the  ratio  of  base  to  height,  0.35  was  given,  but  the  wall 
was  probably  6  inches  in  height  and  it  was  backed  by  sand: 
w  =  89,  w'  =  28,  <p  =  33°  42',  <p'  (assumed)  =  22°,  q  =  —  0.03, 
?o  =  -  0.75. 

The  wall  of  Curie,  Fig.  19,  was  of  wood  coated  on  the  back 
by  sand  with  sand  filling,  so  that  <p  =  <?'  =  35°.  Also  h  = 
0.558  ft,  t  =  0.159  ft.  w/w'  =  3.29,  q  =  0.02,  q0  =  -  1.32. 

It  will  be  observed  in  these  four  walls,  that  R  is  far  outside 
the  base,  indicating  strikingly  the  inaccuracy  of  the  Rankine 
method  when  applied  to  such  walls,  with  the  free  surface  of  the 
earth  horizontal. 

In  the  wall,  Fig.  20,  of  Curie's,  the  retaining  board  AB, 
i  meter  square,  was  coated  with  sand  and  in  the  first  experiment, 
the  angle  BAD  with  the  vertical  AD,  was  27°  30'.  The  "  limit- 


278 


APPENDIX  H 


ing  plane"  (Arts.  27-31),  for  the  earth  surface  BE  level,  bisects 
the  angle  between  the  vertical  and  the  natural  slope  and  thus 
for  the  given  <p  =  33  °  30',  it  makes  an  angle  with  the  vertical 
of  Y2  (90°  -  33°  30')  =  28°  15',  which  is  greater  than  BAD. 
Hence  strictly  the  construction  of  Art.  40,  Fig.  20  applies, 


FIG.  20 

using  <pr  =  <?',  but  as  AB  so  nearly  coincides  with  the  limiting 
plane,  it  is  sufficiently  near  to  use  the  Rankine  method  and 
compute  the  horizontal  earth  thrust  on  the  vertical  plane  AD 
by  the  formula  of  Art.  48  and  combine  this  with  the  weight  of 
sand  BAD  and  the  weight  of  frame  ABC  to  find  the  resultant 
on  the  base  AC.  It  is  found  to  cut  it  o.n  of  its  width  from 
the  outer  toe  C  .".  q  =  o.n. 

In  the  second  experiment,  the  angle  BA  D  was  55  °,  v  =  33  °  30' 
and  AC  =  0.548  meters.  Since  AB  is  below  the  limiting  plane, 
by  Art.  29,  the  same  method  is  pursued,  giving  q  =  0.02.  The 
third  experiment  was  on  a  smaller  retaining  board.  Here 
AB  =  0.2  meter,  BAD  =  55°,  <p  =  <?'  =  35°  and  q  =  0.06. 
Thus  in  the  last  two  experiments,  the  resultant  on  the  base 
passed  practically  through  C. 

In  Fig.  21  is  shown  a  surcharged  wall  of  Curie's,  having 
h  =  2.952  ft.,  t  =  0.755  ft.,  the  level  upper  surface  of  the  sur- 
charge being  4.26  ft.  above  the  top  of  the  wall.  The  surcharge 
extended  over  the  wall  at  the  angle  of  repose  for  the  damp  sand 
filling,  <p  =  45°.  Experiment  gave  <p  =  35°.  The  wall  was  of 
brick  in  Portland  cement  and  w/wr  =  0.62.  The  thrust  on  AF 
was  found  by  the  graphical  method  of  Art.  39  and  it  was  as- 


EXPERIMENTS 


279 


sumed  to  act  (see  Art.  43)  0.364  AF  above  A.  On  combining 
the  thrust  with  the  weight  of  the  wall,  the  resultant  is  found 
to  cut  the  base  0.02  of  its  width  outside  the  base,  so  that  q  = 
—  0.02. 

With  the  exception  of  Trautwine's  wall,  the  filling  was  con- 
tained in  a  box,  so  that  there  was  friction  of  the  filling  on  the 


L 


FIG.  21 


sides  of  the  box,  which  has  been  neglected  in  the  investigation 
above.  When  the  wall  is  long,  this  can  have  but  little  influence 
on  the  results,  but  where  the  length  of  wall  is  not  much  greater 
than  its  height,  the  side  friction  becomes  -appreciable.  By  aid 
of  special  experiments,  Leygue  found  that  for  walls  having  a 
length  twice  the  height,  that  the  true  thrust  was  diminished 
about  5%  from  the  side  friction  for  level-topped  earth  and  as 
much  as  15%  for  the  surface  sloping  at  the  angle  of  repose. 


INDEX 

(The  numbers  refer  to  pages) 
Active  earth  thrust  defined,  29,  56 
Airy,  tensile  strength  of  clay,  8 
Allowable  pressures  on  soils,  n,  125 
Allowable  stresses,  in  concrete,  125,  274 

in  steel  reinforcement,  125,  273 
Angles  of  friction,  Tables,  7-10,  219,  221 
Angles  of  repose,  in  air,  9,  219,  221 

in  water,  10-12 
Angles  of  rupture, 

coherent  earth,  175-181 

non-coherent  earth,  27,  42,  61-64,  IOI 

Bars  of  steel,  bending,  in  concrete,  271 

length  of  embedment,  272 

spacing  of,  270 

working  stresses  in,  125,  273 
Basquin,  Prof.  O.  H., 

circular  diagram  of  stress,  100,  169 

test  of  a  wedge-shaped  glass  beam,  242 
Beams,  reinforced  concrete,  prismatic, 

bond  stress,  268 

diagram  for  R,  259 

formulas  for  resisting  moments,  258 

shearing  stress,  268 
Beams,  wedge-shaped  reinforced  concrete, 

assumptions  in  theory,  241-242 

bond  stress,  265  ^ 

comparison  of  bending  stresses  by  two  methods,  261 

diagonal  tension,  shear  as  a  measure  of,  274 

diagram  for  k  and  j,  255 

resisting  moment  of  concrete,  250,  254,  257 

resisting  moment  of  steel,  249,  254,  257 

shearing  stress,  262-270 

working  stresses,  125,  274 
Bell,  friction  and  cohesion  in  clay,  9 

foundations,  formulas,  195 
Bins,  deep  grain,  theory,  218 

circular  and  square,  220 

experiments  on,  219 

281 


282  INDEX 

(The  numbers  refer  to  pages) 
Bins,  shallow,  coal  heaped,  222 

shallow,  coal  level  at  top,  229 

unsymmetrical,  232 
Bond  stress,  265 

Boussinesq,  modification  of  Rankine  theory,  46,  65 
Bouyant  effort  of  water  in  saturated  earth,  10 
Braced  trenches,  206 
Brunei,  thin  retaining  wall,  206 

Center  of  gravity,  trapezoid,  48 

trapezium,  49 
Center  of  pressure,  coherent  earth,  171,  179,  183 

battered  wall,  197,  205 

braced  trench,  207 

leaning  wall,  205 

surcharged  wall,  208 

uniform  load  on  earth,  187 
Center  of  pressure,  non-coherent  earth,  34,  55,  65,  74 

load  on  earth  surface,  75 

surcharged  walls,  67 
Charmes  reservoir  dam,  7,  191,  198 
Circular  diagram  of  stress,  97,  169 
Clay,  properties,  1-13,  124,  186 
Coal,  angle  of  repose  and  weight,  9,  221 

coefficients  of  friction,  10,  221 

hopper  bins,  222—237 
Coefficients  of  friction,  of  earth,  etc.,  7-9,  221 

earth  and  coal  on  masonry,  iron  and  wood,  10,  221 

grain  on  iron,  wood  and  concrete,  219 
Cohesion,  apparatus  to  determine,  4 

coefficients  of,  7-9 

effect  on  experimental  walls,  3 

experiments  to  determine,  7-9 

laws  of  friction  and  cohesion,  7-9 
Collin,  experiments  relating  to  cohesion,  7 
Concrete,  allowable  stresses  in,  125 

beams  (see  Beams) 
Conjugate  stresses,  15,  85,  86,  170 
Coulomb,  laws  of  friction  and  cohesion,  3,  167 

wedge  of  maximum  thrust,  32,  46 
Counterfort,  design  of,  160,  251 
Counterforted  wall,  149-165 
Criticism  of  theory  of  deep  bins,  k  not  constant,  218 

Rankine's  retaining  wall  theory,  46-47,  51-52,  65 

shallow  bin  theory,  221,  225-229,  231,  233 

stability  of  slopes,  188,  192 

wall  friction  method,  28,  45-47,  72 


INDEX  283 


(The  numbers  refer  to  pages) 
Design  of  retaining  walls,  104-166 

base  slab,  129 

counterforted  wall,  149 

face  slab,  137,  150 

heel,  140,  145,  156,  252,  266 

general  formulas,  overturning,  108-110 

preliminary  design,  133 

requirements,  104 

slab  (gravity)  wall,  127 

tables  of  thicknesses  of  trapezoidal  walls,  116-122 

tee  (T)  wall,  135,  145 

tentative  graphical  method,  52,  106 

toe,  142,  155,  267 

Diagonal  tension,  shear  as  a  measure  of,  274 
Diagrams  for  reinforced  concrete  beams, 

k  and  j  for  wedge-shaped,  255 

k,  j  and  R  for  rectangular,  259 
Direction  of  earth  pressure, 

against  a  battered  wall,  16,  41 

by  Rankine's  method,  50-52,  85,  90,  94,  196 

in  an  unlimited  mass,  14,  90-94 

leaning  (overhanging)  wall,  95 
Dock  walls,  76-79 
Drainage  of  filling,  2,  13,  20,  126,  166,  198 

Earth,  kinds  for  filling,  I,  2,  20,  78,  124 
Earth  foundations,  heaving,  193,  194 
Earth  pressure,  coherent  earth, 

active  and  passive  thrusts,  170 

against  retaining  walls,  196,  208 

center  of,  171,  179,  183,  205,  208 

depth  at  which  pressure  is  zero,  174 

general  graphical  method,  200 

greatest  depth,  unlimited  slope,  173 

maximum  surface  slope  for  given  depth,  171 

Mohr's  circular  diagram  of  stress,  169 

surcharged  walls,  208 

surface  horizontal,  active  thrust,  182-187 

surface  horizontal,  passive  thrust,  192 

surface  of  rupture,  175-181 

uniform  load  on  earth,  187 
Earth  pressure,  non-coherent  earth, 

active  thrust,  29,  56,  59,  69,  89 

center  of,  34,  55,  65,  74 

conjugate  stresses,  15,  85,  86 

Coulomb's  wedge  of  maximum  thrust,  32,  46 

direction  of  (see  Direction  of  pressure) 


INDEX 

(The  numbers  refer  to  pages) 

Earth  pressure,  non-coherent  earth,  ellipse  of  stress,  80-97 

free  surface  level,  64,  70,  89 

free  surface  at  angle  of  repose,  62,  71,  89,  93 

general  formulas,  69,  72,  94,  97 

graphical  methods,  30,  35,  43,  60-64,  87,  90-97 
limiting  plane,  38-42 

Mohr's  circular  diagram  of  stress,  97 

passive  thrust,  30,  53,  79,  89 

planes  of  rupture,  27,  42,  91,  101 

Rankine's  formula,  72,  89 

Rankine's  theorem  as  to  friction,  6 

Rankine's  theory  as  to  retaining  walls,  50-52,  105,  127 

sliding-wedge  theory,  32,  46,  53,  56 

surcharged  wall,  43,  56,  67,  121,  146 

uniform  load  on  earth,  74,  127-165 
Ellipse  of  stress,  80 

applications,  86-97,  169,  243 
Embedment  of  bars,  272 
Expansion  joints,  126,  166 
Experiments,  grain  pressures  in  bins,  219 

retaining  walls,  18,  276 

Factor  of  safety  of  retaining  walls, 

overturning,  25,  104 

sliding,  25,  105,  no,  132,  136,  154 
Filling  behind  retaining  walls,  2,  20,  78,  124 
Flamant,  light-retaining  box,  18 
Footing  of  retaining  wall,  129 
Foundations,  heaving  of  earth,  193 

heaving  of  wall,  194 
Friction,  angle  of,  etc.,  3-10,  221 

coefficients  for  grain,  219 


Gates,  bin,  where  to  place,  220 
Grain,  coefficients  of  friction,  etc.,  219 
Grain  bins,  pressures,  218 

experiments  on,  219 
Graphical  solutions  (see  Earth  pressure} 

pressures  on  shallow  bins,  232 
Gravity  walls,  95,  132 

Heaving  (see  Foundations} 
Heel  slab,  140 

with  fillet,  145,  252,  266 
Historical  notes,  46,  64 
Hopper  bins  (see  Bins} 


INDEX  285 

(The  numbers  refer  to  pages) 

Jacquinot  and  Frontard, 

experiments  relating  to  friction  and  cohesion,  7 
Jamieson,  experiments  on  pressures  on  bins,  219 
Janssen,  experiments  on  pressures  on  bins,  219 

formula,  pressure  on  bins,  218 
Joints,  expansion,  126,  166 

Ketchum's  experiments  on  grain  bins,  219 
theory  for  shallow  bins,  228 

Lateral  pressure  (see  Earth  pressure} 

grain  in  bins,  218 

on  tunnel  linings,  208 

Leygue,  experiments,  friction,  and  cohesion,  7 
Limiting  plane,  38 
Lindenthal's  novel  retaining  wall,  199 

Masonry  (see  Design  of  retaining  walls) 

coefficients  of  friction  and  weights,  10 
Meem,  J.  C.,  pressure  through  saturated  sand,  12 

trenches  and  tunnels,  207,  216 
Mickleton  tunnel,  retaining  wall,  206 
Mohr's  circular  diagram  of  stress,  97,  169 
Moment  of  resistance  of  concrete,  250,  254,  257 

steel,  249,  254,  257 

Neutral  axis,  reinforced  beams,  248,  254,  257,  258 

Obliquity  (see  Direction  of  pressure) 
Overturning  of  retaining  walls,  25,  104,  106 
formulas,  108-115 

Passive  thrust  (see  Earth  pressure) 

Planes  of  rupture  (see  Earth  pressure) 

Pleissner's  experiments,  pressures  on  bins,  219 

Poncelet,  earth  pressure,  64 

Prelini,  construction  for  earth  thrust,  63 

Pressure  (see  Bins,  Braced  Trenches,  Earth  Pressure,  Tunnels) 

Pressure  of  fluids,  71 

Projection  under  base  slab,  144,  158 

Rankine's  theories  (see  Earth  pressure  and  Criticism  of  theory) 
Rebhan,  earth  pressure,  59,  63,  64 
Rectangular  (prismatic)  beams,  258,  268 
Reinforced  concrete  (see  Allowable  stresses,  also  Beams) 
Resal,  assumption  of  uniformly  increasing  pressure  on  any  plane,  47 
obliquity  of  thrust  on  leaning  wall,  95 


286  INDEX 

(The  numbers  refer  to  pages) 

Resal,  planes  of  rupture,  103 

surfaces  of  rupture  in  coherent  earth,  179 

wall  friction  considered,  46 
Retaining  Walls  (see  Design  of) 

expansion  joints,  126,  166 

experiments  on,  18,  276 

weep-holes,  2,  126,  166 
Rods  (see  Bars  of  steel) 
Rupture,  surface  of  (see  Earth  pressure) 

Safe  bearing  capacity  of  soils,  n,  125 
Sand,  properties,  I,  9,  12 
Saturated  earth,  u,  12 

dock  walls,  77 

tunnels  in,  215 

Screw  pile,  supporting  power,  195 
Shear  in  beams,  262-270 
Sliding  of  earth  on  rock  base,  13,  180 
Sliding  of  retaining  wall,  25,  105,  no,  132,  136,  154 
Sliding- wedge  theory,  29-33,  46,  53,  56 
Soil  pressures  allowable,  n,  125 
Soil  reactions  at  base  of  wall,  21-25 
Stability  of  retaining  walls  (see  Overturning  and  Sliding) 
Steel,  allowable  stresses  in,  125,  273  (see  Bars  of  steel) 
Stepping  the  back  of  a  wall  advised,  21,  166 
Stone-retaining  walls,  116-122 
Stress  (see  Beams) 

Stresses  on  a  horizontal  section  of  wall,  21-25 
Surcharge  load,  74,  127,  135,  145,  149,  187 
Surcharged  wall,  43,  56,  67,  121,  146,  208 
Surface  of  rupture  (see  Earth  pressure) 

Tables,  coefficients,  K  and  K\,  34,  42 

coefficients  of  friction,  weights,  etc.,  7-10,  219,  221 

safe  soil  pressures,  1 1 

thicknesses  of  retaining  walls,  116-122 
Tee  (T)  walls,  133-148 
Temperature  reinforcement,  126,  143 
Tentative  design  of  a  retaining  wall,  52-53,  106 
Thrust  (see  Pressure) 
Toe-beam,  142,  155,  267 

Transmission  of  pressure  through  saturated  earth,  12,  76 
Trench,  sloping  face,  stability  of,  187 

vertical  face,  braced,  206 
Tunnels,  pressure  on  bracing,  208 

table  of  pressures  on,  212-214 

saturated  earth,  215 


INDEX  287 


(The  numbers  refer  to  pages) 
Unit  earth  pressure  at  given  depth,  73,  89 

coherent  earth,  170,  171,  176,  179,  180,  182,  192 
Unit  stresses  (see  Beams,  Stresses') 

Vertical  rods  in  counterfort,  159 
Vibration,  21 

Water,  buoyant  effort  on  saturated  earth,  10  (see  Drainage) 

influence  on  thrust,  2,  13,  19-21,  76 

tunnels  in  saturated  earth,  215 
Wedge  of  rupture,  27,  29,  32,  46,  85,  167,  185 
Wedge-shaped  reinforced  concrete  beams,  239-270 
Weep-holes,  2,  126,  166 
Weights,  specific,  of  materials,  9,  10,  219,  221 

in  water,  10,  1 1 
Weyrauch,  theory,  59,  65 
Working  stresses,  125,  273 


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